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
Topics
About: vectors space
The table of contents of this article
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.
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.
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, \(\langle \bullet, \bullet \rangle\), with the topology induced by the metric induced by the norm induced by the inner product
\(f\): \(: V \to V\)
//
Statements:
(
\(f \in \{\text{ the unitary maps }\}\)
\(\implies\)
\(\forall B = \{b_1, ..., b_d\} \in \{\text{ the orthonormal bases for } V\} \text{ with the } d \times d F \text{ matrix } M \text{ such that } f (b_j) = M^l_j b_l (\forall v = v^j b_j \in V (f (v) = v^j M^l_j b_l) \land M \in \{\text{ the unitary matrices }\})\)
)
\(\land\)
(
\(\exists B = \{b_1, ..., b_d\} \in \{\text{ the orthonormal bases for } V\} \text{ with the } d \times d F \text{ matrix } M \text{ such that } f (b_j) = M^l_j b_l (\forall v = v^j b_j \in V (f (v) = v^j M^l_j b_l) \land M \in \{\text{ the unitary matrices }\})\)
\(\implies\)
\(f \in \{\text{ the unitary maps }\}\)
)
//
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.
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.
