1572: Finite Product of Open Maps Is Open

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description/proof of that finite product of open maps is open

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of product map.
• The reader knows a definition of product topology.
• The reader knows a definition of open map.
• The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
• The reader admits the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.

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Target Context

• The reader will have a description and a proof of the proposition that any finite product of open maps is open.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_{1, j} \vert j \in J\}\): \(T_{1, j} \in \{\text{ the topological spaces }\}\)
\(\{T_{2, j} \vert j \in J\}\): \(T_{2, j} \in \{\text{ the topological spaces }\}\)
\(\{f_j \vert j \in J\}\): \(f_j: T_{1, j} \to T_{2, j} \in \{\text{ the open maps }\}\)
//

Statements:
\(\times_{j \in J} f_j \in \{\text{ the open maps }\}\)
//

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2: Note

This proposition requires that \(J\) is finite, because of the reason mentioned in Proof.

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3: Proof

Whole Strategy: Step 1: take any open subset, \(U \subseteq \times_{j \in J} T_{1, j}\), and see that \(U = \cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}\); Step 2: see that \((\times_{j \in J} f_j) (U) = \cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\).

Step 1:

Let \(U \subseteq \times_{j \in J} T_{1, j}\) be any open subset.

\(U = \cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}\) where \(J'\) is a possibly uncountable index set and \(U_{1, j, j'} \subseteq T_{1, j} \) is open: refer to Note for the definition of product topology.

Step 2:

\((\times_{j \in J} f_j) (U) = (\times_{j \in J} f_j) (\cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}) = \cup_{j' \in J'} (\times_{j \in J} f_j) (\times_{j \in J} U_{1, j, j'})\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets, \(= \cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\), by the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.

As each \(f_j\) is open, each \(f_j (U_{1, j, j'}) \subseteq T_{2, j}\) is open.

So, \(\cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\) is open: refer to Note for the definition of product topology.

Note that this proposition requires \(J\) to be finite, because otherwise, for each fixed \(j'\), while only finite of \(f_j (U_{1, j, j'})\) s were allowed to be not \(T_{2, j}\), which would not be guaranteed: the fact that only finite of \(U_{1, j, j'}\) s were not \(T_{1, j}\) would not guaranteed the requirement, because \(f_j\) s were not necessarily surjective.

So, \(\times_{j \in J} f_j\) is open.

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References

<The previous article in this series | The table of contents of this series |

1571: For Product Map, Image of Product Subset Is Product of Images of Component Subsets Under Component Maps

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description/proof of that for product map, image of product subset is product of images of component subsets under component maps

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of product map.

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Target Context

• The reader will have a description and a proof of the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{1, j} \vert j \in J\}\): \(S_{1, j} \in \{\text{ the sets }\}\)
\(\{S_{2, j} \vert j \in J\}\): \(S_{2, j} \in \{\text{ the sets }\}\)
\(\{f_j \vert j \in J\}\): \(f_j: S_{1, j} \to S_{2, j}\)
\(\{S^`_{1, j} \vert j \in J\}\): \(S^`_{1, j} \subseteq S_{1, j}\)
//

Statements:
\((\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j}) = \times_{j \in J} f_j (S^`_{1, j})\)
//

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2: Note

In the definition of product map, the product map by Description 1 and the product map by Description 2 are not exactly the same, but as the logics of Proof for the 2 definitions are basically the same, Proof does not distinguish them.

As a notation, for any \(s \in \times_{j \in J} S_{l, j}\), \(s^m\) denotes the \(m\)-component of \(s\).

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3: Proof

Whole Strategy: Step 1: see that \((\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j}) \subseteq \times_{j \in J} f_j (S^`_{1, j})\); Step 2: see that \(\times_{j \in J} f_j (S^`_{1, j}) \subseteq (\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j})\); Step 3: conclude the proposition.

Step 1:

Let \(s \in (\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j})\) be any.

There is an \(s' \in \times_{j \in J} S^`_{1, j}\) such that \(s = (\times_{j \in J} f_j) (s')\).

\(s' \in \times_{j \in J} S^`_{1, j}\) means that \(s'^j \in S^`_{1, j}\), so, \(f_j (s'^j) \in f_j (S^`_{1, j})\).

\(s = (\times_{j \in J} f_j) (s') = \times_{j \in J} f_j (s'^j) \subseteq \times_{j \in J} f_j (S^`_{1, j})\).

So, \((\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j}) \subseteq \times_{j \in J} f_j (S^`_{1, j})\).

Step 2:

Let \(s \in \times_{j \in J} f_j (S^`_{1, j})\) be any.

\(s^j \in f_j (S^`_{1, j})\).

There is an \(s'^j \in S^`_{1, j}\) such that \(s^j = f_j (s'^j)\).

\(\times_{j \in J} s'^j \in \times_{j \in J} S^`_{1, j}\).

\(s = \times_{j \in J} s^j = (\times_{j \in J} f_j) (\times_{j \in J} s'^j) \in (\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j})\).

So, \(\times_{j \in J} f_j (S^`_{1, j}) \subseteq (\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j})\).

Step 3:

So, \((\times_{j \in J} f_j) (\times_{j \in J} S^`_{1, j}) = \times_{j \in J} f_j (S^`_{1, j})\).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1570: \(C^\infty\) Local Section of \(C^\infty\) Vectors Bundle Is \(C^\infty\) Embedding

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that \(C^\infty\) local section of \(C^\infty\) vectors bundle is \(C^\infty\) embedding

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Topics

About: \(C^\infty\) manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of \(C^\infty\) vectors bundle of rank \(k\).
• The reader knows a definition of open submanifold with boundary of \(C^\infty\) manifold with boundary.
• The reader knows a definition of section of continuous surjection.
• The reader knows a definition of \(C^\infty\) embedding.
• The reader admits the proposition that for any \(C^\infty\) vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at each point on any trivializing open subset.
• The reader admits the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map.
• The reader admits the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases is this.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that any \(C^\infty\) local section of any \(C^\infty\) vectors bundle is a \(C^\infty\) embedding.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
\(k\): \(\in \mathbb{N} \setminus \{0\}\)
\((E, M, \pi)\): \(\in \{\text{ the } C^\infty \text{ vectors bundles of rank } k\}\)
\(U\): \(\in \{\text{ the open embedded submanifolds with boundary of } M\}\)
\(s\): \(: U \to E\), \(\in \{\text{ the } C^\infty \text{ local sections of } \pi\}\)
//

Statements:
\(s \in \{\text{ the } C^\infty \text{ embeddings }\}\)
//

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2: Proof

Whole Strategy: Step 1: see that \(s\) is a \(C^\infty\) injective immersion; Step 2: see that the codomain restriction of \(s\) is a homeomorphism; Step 3: conclude the proposition.

Step 1:

\(s\) is an injection, because for each \(u_1, u_2 \in U\) such that \(u_1 \neq u_2\), \(s (u_1) \neq s (u_2)\), because if \(s (u_1) = s (u_2)\), \(u_1 = \pi \circ s (u_1) = \pi \circ s (u_2) = u_2\), a contradiction against \(u_1 \neq u_2\).

Let us see that \(s\) is a \(C^\infty\) immersion.

\(s\) is \(C^\infty\), by the supposition.

Let \(u \in U\) be any.

Let us take any trivializing chart, \((U_u \subseteq U, \phi_u)\), with a trivialization, \(\Phi: \pi^{-1} (U_u) \to U_u \times \mathbb{R}^k\), by the proposition that for any \(C^\infty\) vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at each point on any trivializing open subset: take a trivializing open neighborhood of \(u\), \(U'_u \subseteq M\), such that \(U'_u \subseteq U\), and take \(U_u\) such that \(U_u \subseteq U'_u\).

By the proposition that for any \(C^\infty\) vectors bundle, the trivialization of any chart trivializing open subset induces the canonical chart map, there is the canonical chart, \((\pi^{-1} (U_u) \subseteq E, \widetilde{\phi_u})\), where \(\widetilde{\phi_u}: \pi^{-1} (U_u) \to U_u \times \mathbb{R}^k \to \mathbb{R}^{d + k} \text{ or } \mathbb{H}^{d + k}, v \mapsto (\pi_2 (\Phi (v)), \phi_u (\pi (v)))\), where \(\pi_2: U_u \times \mathbb{R}^k \to \mathbb{R}^k\) is the projection.

Let us think of the differential of \(s\) at \(u\), \(d s_u: T_uU \to T_{s (u)}E\).

With the standard bases for \(T_uU\) and \(T_{s (u)}E\) with respect to the charts, the components function of \(d s_u\) is \((v^j) \mapsto (\partial_l \hat{s}^j v^l)\) where \(\hat{s} = \widetilde{\phi_u} \circ s \circ {\phi_u}^{-1}: \phi_u (U_u) \subseteq \mathbb{R}^d \to \widetilde{\phi_u} (\pi^{-1} (U_u)) \subseteq \mathbb{R}^{d + k} \text{ or } \mathbb{H}^{d + k}\) is the components function of \(s\), by the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary and any corresponding charts, the components function of the differential of the map with respect to the standard bases is this.

For each \(j \in \{k + 1, ..., k + d\}\), \(\hat{s}^j: (x^1, ..., x^d) \mapsto x^{j - k}\), so, \(\partial_l \hat{s}^j = \delta^{j - k}_l\), so, \(\partial_l \hat{s}^j v^l = \delta^{j - k}_l v^l = v^{j - k}\).

So, for each \(v, v' \in T_uU\) such that \(v \neq v'\), \(v^l \neq v'^l\) for an \(l \in \{1, ..., d\}\), so, \(\partial_l \hat{s}^j v^l = v^{j - k} \neq v'^{j - k} = \partial_l \hat{s}^j v'^l\) for \(j = l + k\).

So, \(d s_u (v) \neq d s_u (v')\).

So, \(d s_u\) is injective, so, \(s\) is a \(C^\infty\) immersion.

Step 2:

Let us see that the codomain restriction of \(s\), \(s': U \to s (U)\), is a homeomorphism.

\(s\) is continuous, because \(s\) is \(C^\infty\), and \(s'\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

As \(s\) is injective, there is the inverse, \(s'^{-1}: s (U) \to U\).

\(s'^{-1}\) is nothing but the restriction of \(\pi\), \(\pi \vert_{s (U)}: s (U) \to U\).

\(\pi\) is continuous, and \(\pi \vert_{s (U)}\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

So, \(s'^{-1}\) is continuous.

So, \(s'\) is a homeomorphism.

Step 3:

So, \(s\) is a \(C^\infty\) embedding.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1569: For Finite-Dimensional Vectors Space with Inner Product with Induced Topology, Map from Space into Space Is Unitary iff Map Is Represented by Unitary Matrix w.r.t. Orthonormal Basis

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description/proof of that for finite-dimensional vectors space with inner product with induced topology, map from space into space is unitary iff map is represented by unitary matrix w.r.t. orthonormal basis

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of unitary map from vectors space with inner product with induced topology into same vectors space.
• The reader knows a definition of dimension of vectors space.
• The reader knows a definition of unitary matrix.
• The reader knows a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
• The reader admits the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_2 \times d_1\) ring matrix.
• The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
• The reader admits the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.
• The reader admits the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.
• The reader admits the proposition that any linear map from any finite-dimensional normed vectors space into any normed vectors space is continuous.
• The reader admits the proposition that any linear map between any vectors metric spaces induced by any norms is continuous if and only if it is bounded.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
• The reader admits the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space with any inner product with the induced topology and any map from the space into the space, the map is represented by the unitary matrix with respect to any orthonormal basis if the map is unitary, and the map is unitary if the map is represented by the unitary matrix with respect to an orthonormal basis.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\), with the topology induced by the metric induced by the norm induced by the inner product
\(f\): \(: V \to V\)
//

Statements:
(
\(f \in \{\text{ the unitary maps }\}\)
\(\implies\)
\(\forall B = \{b_1, ..., b_d\} \in \{\text{ the orthonormal bases for } V\} \text{ with the } d \times d F \text{ matrix } M \text{ such that } f (b_j) = M^l_j b_l (\forall v = v^j b_j \in V (f (v) = v^j M^l_j b_l) \land M \in \{\text{ the unitary matrices }\})\)
)
\(\land\)
(
\(\exists B = \{b_1, ..., b_d\} \in \{\text{ the orthonormal bases for } V\} \text{ with the } d \times d F \text{ matrix } M \text{ such that } f (b_j) = M^l_j b_l (\forall v = v^j b_j \in V (f (v) = v^j M^l_j b_l) \land M \in \{\text{ the unitary matrices }\})\)
\(\implies\)
\(f \in \{\text{ the unitary maps }\}\)
)
//

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2: Note

We need to choose orthonormal bases (instead of just bases), as is seen in Proof.

In order to conclude that \(f\) is unitary, we only need to check with respect to a single orthonormal basis, while if \(f\) is unitary, it is represented by the unitary matrix with respect to any orthonormal basis.

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3: Proof

Whole Strategy: Step 1: suppose that \(f\) is any unitary map; Step 2: take any orthonormal basis, \(B\), and see that \(f\) is represented by \(M\); Step 3: see that \(M\) is unitary; Step 4: suppose that \(f\) is represented by the unitary matrix with respect to an orthonormal basis, \(B\); Step 5: see that \(f\) is a unitary map.

Step 1:

Let us suppose that \(f\) is any unitary map.

Step 2:

Let us take any orthonormal basis for \(V\), \(B = \{b_1, ..., b_d\}\), which is possible by the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.

\(f\) is linear, by the definition of unitary map.

So, \(f\) is represented by the \(d \times d\) \(F\) matrix with respect to \(B\), \(M\), such that \(f (b_j) = M^l_j b_l\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_2 \times d_1\) ring matrix.

Step 3:

The remaining issue is to see that \(M\) is unitary.

\(f\) has the inverse, \(f^{-1} = f^*\), because \(f\) is unitary: \(V^* = V\), so, \(f^*: V \to V\), and \(f^* \circ f = f \circ f^* = id\), which implies that \(f^* = f^{-1}\).

So, \(f\) is bijective and is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

So, \(f\) maps the basis, \(B\), to a basis, by the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain, and \(M\) is invertible, by the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.

\(f^{-1} \circ f (b_j) = b_j\), but the left hand side is \(f^{-1} (M^l_j b_l) = M^l_j f^{-1} (b_l)\).

So, \({M^{-1}}^j_m b_j = {M^{-1}}^j_m M^l_j f^{-1} (b_l) = \delta^l_m f^{-1} (b_l) = f^{-1} (b_m)\).

As \(f\) is unitary, \(\langle b_l, f (b_j) \rangle = \langle f^* (b_l), b_j \rangle = \langle f^{-1} (b_l), b_j \rangle\), but the left hand side is \(\langle b_l, M^m_j b_m \rangle = \overline{M^m_j} \langle b_l, b_m \rangle = \overline{M^m_j} \delta^l_m\), because \(B\) is orthonormal, \(= \overline{M^l_j}\), while the right hand side is \(\langle {M^{-1}}^m_l b_m, b_j \rangle = {M^{-1}}^m_l \langle b_m, b_j \rangle = {M^{-1}}^m_l \delta^j_m = {M^{-1}}^j_l\).

So, \(\overline{M^l_j} = {M^{-1}}^j_l\), which means that \(M^* = M^{-1}\).

So, \(M\) is unitary.

Note that if \(B\) is not chosen to be orthonormal, \(M\) is not guaranteed to be unitary: let \(B\) be orthogonal with \(\langle b_j, b_j \rangle \neq \langle b_l, b_l \rangle\) for some \(j, l\), which is possible because it is just a matter of taking some scalar multiples of \(b_j\) and \(b_l\), then, the left hand side is \(\overline{M^m_j} \langle b_l, b_m \rangle = \overline{M^l_j} \langle b_l, b_l \rangle\) while the right hand side is \({M^{-1}}^m_l \langle b_m, b_j \rangle = {M^{-1}}^j_l \langle b_j, b_j \rangle\), so, \(\overline{M^l_j} \langle b_l, b_l \rangle = {M^{-1}}^j_l \langle b_j, b_j \rangle\), which implies that \(\overline{M^l_j} \neq {M^{-1}}^j_l\), so, \(M\) is not unitary.

Step 4:

Let us suppose that \(f\) is represented by the unitary matrix with respect to an orthonormal basis, \(B\), such that \(f (b_j) = M^l_j b_l\).

Step 5:

\(f\) is linear, by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_2 \times d_1\) ring matrix.

\(f\) is bounded, by the proposition that any linear map from any finite-dimensional normed vectors space into any normed vectors space is continuous and the proposition that any linear map between any vectors metric spaces induced by any norms is continuous if and only if it is bounded.

\(M\) is invertible, so, \(f (b_j) = M^l_j b_l\) maps the basis, \(B\), to a basis, by the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors and the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

\(f\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.

So, there is the inverse, \(f^{-1}: V \to V\).

\(f^{-1} \circ f (b_j) = b_j\), but the left hand side is \(f^{-1} (M^l_j b_l) = M^l_j f^{-1} (b_l)\).

So, \({M^{-1}}^j_m b_j = {M^{-1}}^j_m M^l_j f^{-1} (b_l) = \delta^l_m f^{-1} (b_l) = f^{-1} (b_m)\).

For each \(v = v^j b_j, v' = v'^l b_l \in V\), \(\langle v', f (v) \rangle = \langle v'^l b_l, f (v^j b_j) \rangle = v'^l \langle b_l, v^j f (b_j) \rangle = v'^l \langle b_l, v^j M^m_j b_m \rangle = v'^l \overline{v^j} \overline{M^m_j} \langle b_l, b_m \rangle = v'^l \overline{v^j} \overline{M^m_j} \delta^m_l = v'^l \overline{v^j} \overline{M^l_j} = v'^l \overline{v^j} {M^*}^j_l\).

On the other hand, \(\langle f^{-1} (v'), v \rangle = \langle f^{-1} (v'^l b_l), v^j b_j \rangle = \overline{v^j} \langle v'^l f^{-1} (b_l), b_j \rangle = \overline{v^j} \langle v'^l {M^{-1}}^m_l b_m, b_j \rangle = v'^l \overline{v^j} {M^{-1}}^m_l \langle b_m, b_j \rangle = v'^l \overline{v^j} {M^{-1}}^m_l \delta_{m, j} = v'^l \overline{v^j} {M^{-1}}^j_l\).

As \(M\) is unitary, \(M^* = M^{-1}\), so, \(\langle v', f (v) \rangle = \langle f^{-1} (v'), v \rangle\), which means that \(f^* = f^{-1}\).

So, \(V^* = V\) and \(f^* \circ f = f \circ f^* = id\).

So, \(f\) is unitary.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1568: For Map from Module with \(d_1\)-Elements Basis into Module with \(d_2\)-Elements Basis over Same Ring, Map Is Linear iff Map Is Represented by \(d_2 \times d_1\) Ring Matrix

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description/proof of that for map from module with \(d_1\)-elements basis into module with \(d_2\)-elements basis over same ring, map is linear iff map is represented by \(d_2 \times d_1\) ring matrix

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Topics

About: module
About: matrices space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of basis of module.
• The reader knows a definition of linear map.
• The reader knows a definition of %ring name% matrices space.
• The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

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Target Context

• The reader will have a description and a proof of the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_2 \times d_1\) ring matrix.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\), with any basis, \(B_1 = \{b_{1, 1}, ..., b_{1, d_1}\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\), with any basis, \(B_2 = \{b_{2, 1}, ..., b_{2, d_2}\}\)
\(f\): \(: M_1 \to M_2\)
\(M\): \(\in \{\text{ the } d_2 \times d_1 R \text{ matrices }\}\), such that \(f (b_{1, j}) = M^l_j b_{2, l}\)
//

Statements:
\(f \in \{\text{ the linear maps }\}\)
\(\implies\)
\(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\)
//

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2: Note

\(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\) means nothing but "\(f\) is represented by \(M\)", because \(M\) determines \(f\), because the decomposition, \(v^j b_{1, j}\), is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

When \(R\) is commutative, it means that \(\begin{pmatrix} {v_2}^1 \\ ... \\ {v_2}^{d_2} \end{pmatrix} = M \begin{pmatrix} {v_1}^1 \\ ... \\ {v_1}^{d_1} \end{pmatrix}\) where \(v_2 := f (v_1) = {v_2}^l b_{2, l}\); when \(R\) is not commutative, \({v_1}^j M^l_j b_{2, l}\) cannot be expressed so, because \({v_1}^j M^l_j \neq M^l_j {v_1}^j\) in general while \(M \begin{pmatrix} {v_1}^1 \\ ... \\ {v_1}^{d_1} \end{pmatrix}\) produces the latter.

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3: Proof

Whole Strategy: Step 1: see that \(M\) is well-defined and \({v_1}^j M^l_j b_{2, l}\) is uniquely determined; Step 2: suppose that \(f\) is linear; Step 3: see that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\); Step 4: suppose that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\); Step 5: see that \(f\) is linear.

Step 1:

Let us see that \(M\) is well-defined.

\(M\) is uniquely determined, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: \(M^l_j b_{2, l}\) is the decomposition of \(f (b_{1, j})\) with respect to \(B_2\) where \(M^l_j \in R\).

\(M\) is indeed a \(d_2 \times d_1\) \(R\) matrix, because \(l \in \{1, ..., d_2\}\) and \(j \in \{1, ..., d_1\}\): when an \(l\) does not appear in \(M^l_j b_{2, l}\), \(M^l_j := 0\).

The decomposition, \(v_1 = {v_1}^j b_{1, j}\), is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: when a \(j\) does not appear in \({v_1}^j b_{1, j}\), \({v_1}^j := 0\).

So, \({v_1}^j M^l_j b_{2, l}\) is uniquely determined.

Step 2:

Let us suppose that \(f\) is linear.

Step 3:

Let us see that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\).

\(f (v_1) = f ({v_1}^j b_{1, j}) = {v_1}^j f (b_{1, j})\), because \(f\) is linear, \(= {v_1}^j M^l_j b_{2, l}\).

Step 4:

Let us suppose that \(\forall v_1 = {v_1}^j b_{1, j} \in M_1 (f (v_1) = {v_1}^j M^l_j b_{2, l})\).

Let \(v_1 = {v_1}^j b_{1, j}, v'_1 = {v'_1}^j b_{1, j} \in M_1\) and \(r, r' \in R\) be any.

\(f (r v_1 + r' v'_1) = f (r {v_1}^j b_{1, j} + r' {v'_1}^j b_{1, j}) = f ((r {v_1}^j + r' {v'_1}^j) b_{1, j}) = (r {v_1}^j + r' {v'_1}^j) M^l_j b_{2, l} = r ({v_1}^j M^l_j b_{2, l}) + r' ({v'_1}^j M^l_j b_{2, l}) = r f (v_1) + r' f (v'_1)\), which means that \(f\) is linear.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1567: For Finite-Dimensional Vectors Space with Inner Product, Components Matrix of Inner Product w.r.t. Basis Is Invertible

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description/proof of that for finite-dimensional vectors space with inner product, components matrix of inner product w.r.t. basis is invertible

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of inner product on real or complex vectors space.
• The reader knows a definition of %ring name% matrices space.
• The reader knows a definition of determinant of square matrix over ring.
• The reader knows a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
• The reader admits the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
• The reader admits the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space with any inner product, the components matrix of the inner product with respect to any basis is invertible.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, ..., b_d\}\)
\(M\): \(\in \{\text{ the } d \times d F \text{ matrices }\}\), such that \(M^j_l = \langle b_l, b_j \rangle\)
//

Statements:
\(det M \neq 0\)
//

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2: Proof

Whole Strategy: Step 1: take the Gram-Schmidt orthonormalization of \(B\), \(B' = \{b'_1, ..., b'_d\}\), and see that \(M'\) with respect to \(B'\) is \(I\); Step 2: see that \(b_j = b'_l N^l_j\) where \(det N \neq 0\); Step 3: see that \(M = N^* M' N\); Step 4: see that \(det M = det (N^* M' N) \neq 0\).

Step 1:

Let us take the Gram-Schmidt orthonormalization of \(B\), \(B' = \{b'_1, ..., b'_d\}\).

\(B'\) is a basis for \(V\), by the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent and the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

Let \(M'\) be the \(d \times d\) \(F\) matrix such that \(M'^j_l = \langle b'_l, b'_j \rangle\).

\(M' = I\), because \(B'\) is orthonormal.

Step 2:

For each \(j \in \{1, ..., d\}\), \(b_j = b'_l N^l_j\) where \(N^l_j \in F\), because \(B'\) is a basis.

\(det N \neq 0\), by the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.

Step 3:

\(M^j_l = \langle b_l, b_j \rangle = \langle b'_m N^m_l, b'_n N^n_j \rangle = \overline{N^n_j} \langle b'_m, b'_n \rangle N^m_l = \overline{N^n_j} M'^n_m N^m_l = (N^* M' N)^j_l\), so, \(M = N^* M' N\).

Step 4:

\(det M = det (N^* M' N) = det N^* det M' det N \neq 0\), because \(det N^* = \overline{det N} \neq 0\), \(det M' = det I = 1 \neq 0\), and \(det N \neq 0\).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1566: Inverse of Unitary Matrix Is Unitary

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description/proof of that inverse of unitary matrix is unitary

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Topics

About: matrices space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of unitary matrix.
• The reader admits the proposition that the Hermitian conjugate of any unitary matrix is unitary.

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Target Context

• The reader will have a description and a proof of the proposition that the inverse of any unitary matrix is unitary.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(M\): \(\in \{\text{ the unitary matrices }\}\)
\(M^{-1}\): \(= \text{ the inverse of } M\)
//

Statements:
\(M^{-1} \in \{\text{ the unitary matrices }\}\)
//

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2: Proof

Whole Strategy: Step 1: see that \(M^{-1} = M^*\); Step 2: apply the proposition that the Hermitian conjugate of any unitary matrix is unitary.

Step 1:

\(M^{-1} = M^*\), by the definition of unitary matrix.

Step 2:

\(M^*\) is unitary, by the proposition that the Hermitian conjugate of any unitary matrix is unitary, so, \(M^{-1}\) is unitary.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1565: Hermitian Conjugate of Unitary Matrix Is Unitary

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that Hermitian conjugate of unitary matrix is unitary

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Topics

About: matrices space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of unitary matrix.
• The reader knows a definition of Hermitian conjugate of complex matrix.
• The reader admits the proposition that the Hermitian conjugate of the Hermitian conjugate of any complex matrix is the matrix.

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Target Context

• The reader will have a description and a proof of the proposition that the Hermitian conjugate of any unitary matrix is unitary.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(M\): \(\in \{\text{ the unitary matrices }\}\)
\(M^*\): \(= \text{ the Hermitian conjugate of } M\)
//

Statements:
\(M^* \in \{\text{ the unitary matrices }\}\)
//

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2: Proof

Whole Strategy: Step 1: see that \({M^*}^* = M\); Step 2: see that \({M^*}^* = {M^*}^{-1}\).

Step 1:

\({M^*}^* = M\), by the proposition that the Hermitian conjugate of the Hermitian conjugate of any complex matrix is the matrix.

Step 2:

\({M^*}^* M^* = M M^* = I\), because \(M\) is unitary.

\(M^* {M^*}^* = M^* M = I\), because \(M\) is unitary.

So, \({M^*}^* = {M^*}^{-1}\).

So, \(M^*\) is unitary.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1564: Hermitian Conjugate of Hermitian Conjugate of Complex Matrix Is Matrix

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that Hermitian conjugate of Hermitian conjugate of complex matrix is matrix

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Topics

About: matrices space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of Hermitian conjugate of complex matrix.

---

Target Context

• The reader will have a description and a proof of the proposition that the Hermitian conjugate of the Hermitian conjugate of any complex matrix is the matrix.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(M\): \(\in \{\text{ the } m \times n \text{ complex matrices }\}\)
\({M^*}^*\): \(= \text{ the Hermitian conjugate of the Hermitian conjugate of } M\)
//

Statements:
\({M^*}^* = M\)
//

---

2: Proof

Whole Strategy: Step 1: see that \({M^*}^*\) is an \(m \times n\) matrix; Step 2: see that \({{M^*}^*}^j_l = M^j_l\).

Step 1:

\(M^*\) is an \(n \times m\) matrix.

\({M^*}^*\) is an \(m \times n\) matrix.

Step 2:

\({{M^*}^*}^j_l = \overline{{M^*}^l_j} = \overline{\overline{M^j_l}} = M^j_l\).

So, \({M^*}^* = M\).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1563: For Map Between Sets, Image of Intersection of Preimages of Codomain Subsets Is Intersection of Images of Preimages of Codomain Subsets

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description/proof of that for map between sets, image of intersection of preimages of codomain subsets is intersection of images of preimages of codomain subsets

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Topics

About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of map.
• The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.

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Target Context

• The reader will have a description and a proof of the proposition that for any map between any sets, the image of the intersection of the preimages of any codomain subsets is the intersection of the images of the preimages of the codomain subsets.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{2, j} \subseteq S_2 \vert j \in J\}\):
//

Statements:
\(f (\cap_{j \in J} f^{-1} (S_{2, j})) = \cap_{j \in J} f (f^{-1} (S_{2, j}))\)
//

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2: Proof

Whole Strategy: Step 1: see that \(f (\cap_{j \in J} f^{-1} (S_{2, j})) \subseteq \cap_{j \in J} f (f^{-1} (S_{2, j}))\); Step 2: see that \(\cap_{j \in J} f (f^{-1} (S_{2, j})) \subseteq f (\cap_{j \in J} f^{-1} (S_{2, j}))\); Step 3: conclude the proposition.

Step 1:

Let \(s \in f (\cap_{j \in J} f^{-1} (S_{2, j}))\) be any.

There is an \(s' \in \cap_{j \in J} f^{-1} (S_{2, j})\) such that \(f (s') = s\).

\(s' \in f^{-1} (S_{2, j})\) for each \(j \in J\).

\(s = f (s') \in f (f^{-1} (S_{2, j}))\) for each \(j\).

So, \(s \in \cap_{j \in J} f (f^{-1} (S_{2, j}))\).

So, \(f (\cap_{j \in J} f^{-1} (S_{2, j})) \subseteq \cap_{j \in J} f (f^{-1} (S_{2, j}))\).

Step 2:

Let \(s \in \cap_{j \in J} f (f^{-1} (S_{2, j}))\) be any.

\(s \in f (f^{-1} (S_{2, j}))\) for each \(j \in J\).

Let \(l \in J\) be any fixed one.

As \(s \in f (f^{-1} (S_{2, l}))\), there is an \(s' \in f^{-1} (S_{2, l})\) such that \(f (s') = s\).

Let \(m \in J\) be any.

\(s = f (s') \in f (f^{-1} (S_{2, m}))\).

So, \(s' \in f^{-1} (f (f^{-1} (S_{2, m})))\).

But \(f (f^{-1} (S_{2, m})) \subseteq S_{2, m}\).

So, \(f^{-1} (f (f^{-1} (S_{2, m}))) \subseteq f^{-1} (S_{2, m})\).

\(s' \in f^{-1} (f (f^{-1} (S_{2, m}))) \subseteq f^{-1} (S_{2, m})\).

So, \(s' \in \cap_{j \in J} f^{-1} (S_{2, j})\).

So, \(s = f (s') \in f (\cap_{j \in J} f^{-1} (S_{2, j}))\).

So, \(\cap_{j \in J} f (f^{-1} (S_{2, j})) \subseteq f (\cap_{j \in J} f^{-1} (S_{2, j}))\).

Step 3:

So, \(f (\cap_{j \in J} f^{-1} (S_{2, j})) = \cap_{j \in J} f (f^{-1} (S_{2, j}))\).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1562: For Map Between Sets, Image of Preimage of Codomain Subset Is Intersection of Codomain Subset and Map Range

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for map between sets, image of preimage of codomain subset is intersection of codomain subset and map range

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Topics

About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of map.
• The reader admits the proposition that for any map between any sets, the image of the intersection of the intersection of the preimages of any codomain subsets and any domain subset is the intersection of the intersection of the codomain subsets and the image of the domain subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any map between any sets, the image of the preimage of any codomain subset is the intersection of the codomain subset and the map range.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(f\): \(: S_1 \to S_2\)
\(S^`_2\): \(\subset S_2\)
//

Statements:
\(f (f^{-1} (S^`_2)) = S^`_2 \cap f (S_1)\)
//

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2: Proof

Whole Strategy: Step 1: apply the proposition that for any map between any sets, the image of the intersection of the intersection of the preimages of any codomain subsets and any domain subset is the intersection of the intersection of the codomain subsets and the image of the domain subset.

Step 1:

For the proposition that for any map between any sets, the image of the intersection of the intersection of the preimages of any codomain subsets and any domain subset is the intersection of the intersection of the codomain subsets and the image of the domain subset, \(\cap_{j \in J} f^{-1} (S_{2, j})\) can be taken to be \(f^{-1} (S^`_2)\) and the domain subset can be taken to be \(S_1\), then \(f (f^{-1} (S^`_2) \cap S_1) = S^`_2 \cap f (S_1)\), but \(f (f^{-1} (S^`_2) \cap S_1) = f (f^{-1} (S^`_2))\).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>