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definition of left-invariant vectors field over Lie group
Topics
About:
group
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of left-invariant vectors field over Lie group.
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:
\( G\): \(\in \{\text{ the Lie groups }\}\)
\(*V\): \(\in \{\text{ the vectors fields over } G\}\)
//
Conditions:
\(\forall g \in G ((V, V) \in \{\text{ the } l_g \text{ -related vectors fields pairs }\})\), where \(l_g: G \to G, g' \mapsto g g'\) is the left-translation by \(g\)
//
2: Note
\(V\) is inevitably \(C^\infty\), by the proposition that any left-invariant vectors field over any Lie group is \(C^\infty\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and subspace with some connected-components removed, remained connected-components are connected-components of subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and the subspace with any some connected-components removed, the remained connected-components are the connected-components of the subspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S'\): \(= \{C_j \in \{\text{ the connected components of } T'\} \vert j \in J\}\)
\(S\): \(\subset S'\)
\(T\): \(= \cup S \subset T'\), with the subspace topology
//
Statements:
\(S = \{\text{ the connected components of } T\}\)
//
2: Proof
Whole Strategy: Step 1: see that each \(C_j \in S\) is connected on \(T\); Step 2: see that for each \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\); Step 3: conclude the proposition.
Step 1:
Each \(C_j \in S\) is connected on \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Step 2:
Let us see that for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Let us suppose that a \(c_j \in C_j\) was connected to a \(c_l \in C_l\) on \(T\).
There would be a connected subspace of \(T\), \(T^` \subseteq T\), such that \(c_j, c_l \in T^`\).
\(T^`\) would be a connected subspace of \(T'\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, which would mean that \(c_j\) and \(c_l\) were connected on \(T'\), a contradiction against that \(c_j\) and \(c_l\) were not in the same connected-component of \(T'\).
So, for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Step 3:
So, each \(C_j\) is a connected subspace of \(T\) that cannot be made larger, so, \(C_j\) is a connected component of \(T\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and set of connected subspaces, subspace is contained in connected component of union of subspaces
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and any set of connected subspaces, each subspace is contained in a connected component of the union of the subspaces.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \vert j \in J\}\): \(T_j \in \{\text{ the connected topological subspaces of } T'\}\)
\(T\): \(= \cup_{j \in J} T_j \subseteq T'\) with the subspace topology
//
Statements:
\(\forall j \in J (\exists C \in \{\text{ the connected components of } T\} (T_j \subseteq C))\)
//
2: Proof
Whole Strategy: Step 1: see that \(T_j\) is a connected subspace of \(T\); Step 2: conclude the proposition.
Step 1:
\(T_j\) is a connected subspace of \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Step 2:
Let us take the union of all the connected subspaces of \(T\) that contain \(T_j\), which is a connected subspace of \(T\) that cannot be made larger, by the proposition that for any topological space, the union of any connected subspaces that share any point is connected: that cannot be made larger because if there was a larger one, the larger one would be contained in the union.
So, by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger, the union is a connected component of \(T\) in which \(T_j\) is contained.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for locally connected topological space, open subspace is locally connected
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any locally connected topological space, any open subspace is locally connected.
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:
\(T'\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(T\): \(\in \{\text{ the open subsets of } T'\}\) with the subspace topology
//
Statements:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
//
2: Note
As Proof shows, \(T\) needs to be open in order for this proposition to be applied.
3: Proof
Whole Strategy: Step 1: for each \(t \in T\), take any neighborhood of \(t\) on \(T\), \(N'_t\), and see that \(N'_t\) is a neighborhood of \(t\) on \(T'\); Step 2: take a connected neighborhood of \(t\) on \(T'\) contained in \(N'_t\), \(N_t\); Step 3: see that \(N_t\) is a connected neighborhood of \(t\) on \(T\).
Step 1:
Let \(t \in T\) be any.
Let \(N'_t \subseteq T\) be any neighborhood of \(t\).
\(N'_t\) is a neighborhood of \(t \in T'\) on \(T'\), because there is an open neighborhood of \(t\) on \(T\), \(U'_t \subseteq T\), such that \(U'_t \subseteq N'_t\), which is open also on \(T'\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Step 2:
There is a connected neighborhood of \(t\) on \(T'\), \(N_t \subseteq T'\), such that \(t \in N_t \subseteq N'_t\), by the definition of locally connected topological space.
Step 3:
\(N_t \subseteq N'_t \subseteq T\) is a neighborhood of \(T\), because there is an open neighborhood of \(t\) on \(T'\), \(U_t \subseteq T'\), such that \(U_t \subseteq N_t\), which is an open neighborhood of \(t\) also on \(T\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
While \(N_t\) is a connected subspace of \(T'\), \(N_t\) is a connected subspace of \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(N_t\) is a connected neighborhood of \(t\) on \(T\).
So, \(T\) is locally connected.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and \(2\) connected subspaces, connected-components of union of subspaces are subspaces or union of subspaces
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and any \(2\) connected subspaces, the connected-components of the union of the subspaces are the subspaces or the union of the subspaces.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T_1\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_2\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_1 \cup T_2\): \(\subseteq T'\) as the topological subspace
//
Statements:
\(T_1 \cup T_2\) has the connected-components, \(\{T_1, T_2\}\) or \(\{T_1 \cup T_2\}\)
//
2: Note
Refer to the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace and the proposition that for any topological space, the union of any connected subspaces that share any point is connected.
3: Proof
Whole Strategy: Step 1: see that for each \(t_1, t'_1 \in T_1\), \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\) and for each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\); Step 2: suppose that each \(t_1\) is not connected to any \(t_2\), and see that \(\{T_1, T_2\}\) are the connected components; Step 3: suppose that a \(t_1\) is connected to a \(t_2\), and see that \(\{T_1 \cup T_2\}\) is the connected component.
Step 1:
Let \(t_1, t'_1 \in T_1\) be any.
\(T_1\) as the subspace of \(T'\) is the subspace of \(T_1 \cup T_2\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(T_1\) is a connected subspace of \(T_1 \cup T_2\) that contains \(t_1\) and \(t'_1\).
So, \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\).
For each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\).
Step 2:
There are only the 2 cases: 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\); 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Let us suppose that 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\).
\(T_1\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), because each pair of points on \(T_1\) are connected by Step 1 and each point not on \(T_1\) is on \(T_2\) and does not belong to the equivalence class, by the supposition.
\(T_2\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), likewise.
So, \(\{T_1, T_2\}\) are the connected components of \(T_1 \cup T_2\).
Step 3:
Let us suppose that 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Each pair of points on \(T_1 \cup T_2\) are connected, because when they are \(t'_1, t''_1 \in T_1\), they are connected by Step 1; when they are \(t'_1 \in T_1\) and \(t'_2 \in T_2\), they are connected because \(t'_1\) and \(t_1\) are connected, \(t_1\) and \(t_2\) are connected, and \(t_2\) and \(t'_2\) are connected; when they are \(t'_2, t''_2 \in T_2\), they are connected by Step 1.
So, \(T_1 \cup T_2\) is the equivalence class of connected-ness on \(T_1 \cup T_2\).
So, \(\{T_1 \cup T_2\}\) is the connected component of \(T_1 \cup T_2\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for set, union of subsets minus subset is union of (each of former subsets minus latter subset) s
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any set, the union of any subsets minus any subset is the union of (each of the former subsets minus the latter subset) s.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S' \vert j \in J\}\):
\(S\): \(\subseteq S'\)
//
Statements:
\((\cup_{j \in J} S_j) \setminus S = \cup_{j \in J} (S_j \setminus S)\)
//
2: Proof
Whole Strategy: Step 1: see that for each \(s \in (\cup_{j \in J} S_j) \setminus S\), \(s \in \cup_{j \in J} (S_j \setminus S)\); Step 2: see that for each \(s \in \cup_{j \in J} (S_j \setminus S)\), \(s \in (\cup_{j \in J} S_j) \setminus S\).
Step 1:
Let \(s \in (\cup_{j \in J} S_j) \setminus S\) be any.
\(s \in \cup_{j \in J} S_j\), so, \(s \in S_j\) for a \(j \in J\).
\(s \notin S\).
So, \(s \in S_j \setminus S\) for that \(j\).
So, \(s \in \cup_{j \in J} (S_j \setminus S)\).
Step 2:
Let \(s \in \cup_{j \in J} (S_j \setminus S)\) be any.
\(s \in S_j \setminus S\) for a \(j \in J\).
So, \(s \in S_j\) for that \(j\) and \(s \notin S\).
So, \(s \in \cup_{j \in J} S_j\).
So, \(s \in (\cup_{j \in J} S_j) \setminus S\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that antisymmetric real matrix can be block-diagonalized by orthogonal matrix
Topics
About:
matrices space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any antisymmetric real matrix can be block-diagonalized by an orthogonal 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:
\(M\): \(\in \{\text{ the } n \times n \text{ real antisymmetric matrices } \}\)
//
Statements:
\(\exists O \in \{\text{ the orthogonal matrices }\} (O^t M O = \begin{pmatrix} 0 & \sqrt{\lambda_1} & 0 & ... & & & & & & & 0 \\ - \sqrt{\lambda_1} & 0 & ... & & & & & & & & 0 \\ 0 & ... & 0 & \sqrt{\lambda_2} & 0 & ... & & & & & 0 \\ 0 & 0 & - \sqrt{\lambda_2} & 0 & ... & & & & & & 0 \\ ... \\ 0 & ... & & && & 0 & \sqrt{\lambda_{2 m}} & 0 & ... & 0 \\ 0 & ... & & & & 0 & - \sqrt{\lambda_{2 m}} & 0 & ... & & 0 \\ 0 & ... & & & & & & & & & 0 \\ ... \\ 0 & ... & & & & & & & & & 0 \end{pmatrix})\), where \(\{\lambda_1, ..., \lambda_{2 m}, 0, ..., 0\}\) are the eigenvalues of \(M^t M\) where \(0 \lt \lambda_j\)
//
2: Note
"block-diagonalized" means that the result has some diagonal blocks (which means blocks at diagonal positions, not blocks having diagonal shapes), each of which is \(\begin{pmatrix} 0 & \sqrt{\lambda_j} \\ - \sqrt{\lambda_j} & 0 \end{pmatrix}\) or \(\begin{pmatrix} 0 \end{pmatrix}\), with the other components \(0\).
Obviously, any nonzero antisymmetric matrix cannot be diagonalized, because the diagonal components of the antisymmetric matrix are all \(0\), so, \(O^t M O\) would be the \(0\) matrix (\(O^t M O\) is antisymmetric), and \(M = O O^t M O O^t = 0\).
3: Proof
Whole Strategy: Step 1: see that \(M^t M\) is symmetric and has the decreasing eigenvalues (with any duplications), \((\lambda_1, ..., \lambda_k, 0, ..., 0)\), where \(0 \lt \lambda_j\), with some eigenvectors, \(e_1, ..., e_n\); Step 2: see that \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\), so, \((e_j, M e_j)\) is a pair for the same \(\lambda_j\); Step 3: take an orthonormal eigenvectors of \(M^t M\), \((O_1, ..., O_{2 m}, O_{2 m + 1}, ..., O_n)\), with the eigenvalues, \((\lambda_1, ..., \lambda_{2 m}, 0, ..., 0)\); Step 4: take \(O\) as \(\begin{pmatrix} O_1 & ... & O_n \end{pmatrix}\); Step 5: see that \(O^t M O\) is as is demanded.
Step 1:
\(M^t M\) is symmetric, because \((M^t M)^t = M^t {M^t}^t\), by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, \(= M^t M\).
So, \(M^t M\) has the eigenvalues (ordered decreasingly for our convenience), \((\lambda_1, ..., \lambda_n)\), with any duplications, with some eigenvectors, \((e_1, ..., e_n)\), as is well known.
Let us see that \(0 \le \lambda_j\) for each \(j \in \{1, ..., n\}\).
\(M^t M e_j = \lambda_j e_j\).
\({e_j}^t M^t M e_j = (M e_j)^t M e_j = \Vert M e_j \Vert^2\), which is non-negative.
But \({e_j}^t M^t M e_j = {e_j}^t \lambda_j e_j = \lambda_j {e_j}^t e_j = \lambda_j \Vert e_j \Vert^2\).
So, \(0 \le \lambda_j \Vert e_j \Vert^2\), which implies that \(0 \le \lambda_j\).
So, \((\lambda_1, ..., \lambda_n) = (\lambda_1, ..., \lambda_k, 0, ..., 0)\) where \(0 \lt \lambda_j\), where the "\(0, ..., 0\)" part does not really exist when \(k = n\).
Step 2:
Let \(j \in \{1, ..., k\}\) be any.
Let us see that \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\).
\(M^t M (M e_j) = - M^t (- M) (M e_j) = - (- M) (M^t) (M e_j) = M (M^t M e_j) = M (\lambda_j e_j) = \lambda_j (M e_j)\).
On the other hand, \(M (M e_j) = - (- M M e_j) = - (M^t M e_j) = - \lambda_j e_j \neq 0\), which implies that \(M e_j \neq 0\).
\(e_j = - 1 / \lambda_j M (M e_j)\).
\({e_j}^t (M e_j) = (- 1 / \lambda_j M (M e_j))^t M e_j = - 1 / \lambda_j (M (M e_j))^t M e_j = - 1 / \lambda_j (M e_j)^t M^t M e_j = - 1 / \lambda_j (M e_j)^t \lambda_j e_j = - (M e_j)^t e_j = - ((M e_j)^t e_j)^t\), because the transpose of any scalar is the scalar, \(= - {e_j}^t ((M e_j)^t)^t = - {e_j}^t (M e_j)\), which implies that \({e_j}^t (M e_j) = 0\).
So, \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\).
So, \(\{e_j, M e_j\}\) is linearly independent, and \((e_j, M e_j)\) forms a pair of eigenvectors for \(\lambda_j\).
Step 3:
\(O_1 := e_1 / \Vert e_1 \Vert\) is a normal eigenvector for \(\lambda_1\).
Let us take \(O_2 := - 1 / \sqrt{\lambda_1} M O_1\), which is an eigenvector for \(\lambda_1\) orthogonal to \(O_1\) by Step 2.
\({O_2}^t O_2 = (- 1 / \sqrt{\lambda_1} M O_1)^t (- 1 / \sqrt{\lambda_1} M O_1) = 1 / \lambda_1 (M O_1)^t M O_1 = 1 / \lambda_1 {O_1}^t M^t M O_1 = 1 / \lambda_1 {O_1}^t \lambda_1 O_1 = {O_1}^t O_1 = 1\), so, \(O_2\) is a normal eigenvector for \(\lambda_1\) orthogonal to \(O_1\).
Note that \(M O_2 = M (- 1 / \sqrt{\lambda_1} M O_1) = - 1 / \sqrt{\lambda_1} M M O_1 = \sqrt{\lambda_1} O_1\), by Step 2.
If there is no more duplication of \(\lambda_1\), \((\lambda_1, \lambda_2 = \lambda_1)\) will be the duplications of \(\lambda_1\).
Let us suppose that there is another duplication of \(\lambda_1\).
A normal eigenvector, \(O_3\), can be taken to be orthogonal to \((O_1, O_2)\), by the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
Then, let us take \(O_4 := - 1 / \sqrt{\lambda_1} M O_3\), a normal eigenvector for \(\lambda_1\) orthogonal to \(O_3\), as before.
Let us see that \(O_4\) is orthogonal also to \(O_1\) and \(O_2\).
For \(j \in \{1, 2\}\), \({O_j}^t O_4 = {O_j}^t (- 1 / \sqrt{\lambda_1} M O_3) = - 1 / \sqrt{\lambda_1} {O_j}^t M O_3 = - 1 / \sqrt{\lambda_1} {O_j}^t {M^t}^t O_3 = - 1 / \sqrt{\lambda_1} (M^t O_j)^t O_3 = - 1 / \sqrt{\lambda_1} (- M O_j)^t O_3 = 1 / \sqrt{\lambda_1} (M O_j)^t O_3\), but \(M O_j\) is a scalar multiple of \(O_1\) or \(O_2\), so, \(= 0\).
And so on, after all, \(\lambda_1\) has some even duplications, \((\lambda_1, \lambda_2 = \lambda_1, ..., \lambda_{2 l - 1} = \lambda_1, \lambda_{2 l} = \lambda_1)\), with the orthonormal eigenvectors, \((O_1, O_2, ..., O_{2 l - 1}, O_{2 l})\).
Doing likewise for each eigenvalue-positive-duplications, we have the eigenvalues, \((\lambda_1, ..., \lambda_{2 m})\) with the orthonormal eigenvectors, \((O_1, ..., O_{2 m})\): any 2 eigenvectors with different eigenvalues, \(O_j, O_l\), are inevitably orthogonal to each other, because \((\lambda_l - \lambda_j) {O_j}^t O_l = \lambda_l {O_j}^t O_l - \lambda_j {O_j}^t O_l = {O_j}^t M^t M O_l - (M^t M O_j)^t O_l = ((M^t M)^t O_j)^t O_l - (M^t M O_j)^t O_l = (M^t M O_j)^t O_l - (M^t M O_j)^t O_l = 0\), which implies that \({O_j}^t O_l = 0\).
For the eigenvalue-0-duplications, we take any orthonormal eigenvectors, by the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
So, we have the eigenvalues \((\lambda_1, ..., \lambda_{2 m}, 0, ..., 0)\) with the orthonormal eigenvectors, \((O_1, ..., O_{2 m}, O_{2 m + 1}, ..., O_n)\): any 2 eigenvectors with different eigenvalues, \(O_j, O_l\), are inevitably orthogonal to each other, as before.
Step 4:
Let us take \(O := \begin{pmatrix} O_1 & ... & O_n \end{pmatrix}\).
\(O\) is an orthogonal matrix, because \((O_1, ..., O_n)\) is orthonormal: \((O_1, ..., O_n)\)'s being orthonormal is nothing but \(O^t O = I\).
Step 5:
Let us see that \(O^t M O\) is as is demanded.
For each \(2 m \lt j\), \(M O_j = 0\), because \(M^t M O_j = 0\), so, \({O_j}^t M^t M O_j = 0\), but the left hand side is \((M O_j)^t M O_j = \Vert M O_j \Vert^2\), so, \(\Vert M O_j \Vert^2 = 0\), which implies that \(M O_j = 0\).
Let us see that \((O^t M O)^j_l = {O_j}^t M O_l\).
\((O^t M O)^j_l = (O^t)^j (M O)_l\), where \((O^t)^j\) denotes the \(j\)-th row of \(O^t\) and \((M O)_l\) denotes the \(l\)-th column of \(M O\).
\((O^t)^j = {O_j}^t\).
\((M O)_l = M O_l\).
So, \((O^t M O)^j_l = {O_j}^t M O_l\).
For each \(j = 2 r + 1\) for each \(r \in \{0, ..., m - 1\}\), for \(l = j + 1\), \({O_j}^t M O_l = {O_j}^t \sqrt{\lambda_j} O_j = \sqrt{\lambda_j}\), and for any other \(l\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
For each \(j = 2 r + 2\) for each \(r \in \{0, ..., m - 1\}\), for \(l = j - 1\), \({O_j}^t M O_l = {O_j}^t (- \sqrt{\lambda_j} O_j) = - \sqrt{\lambda_j}\), and for any other \(l\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
For each \(j\) such that \(2 m \lt j\), for each \(l \in \{1, .., n\}\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
That meas that \(O^t M O\) is as is demanded.
References
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description/proof of that for continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology, if preimage of nonzeros is contained in compact subset of domain, map is uniformly-continuous
Topics
About:
group
About:
topological space
The table of contents of this article
Starting Context
-
The reader knows a definition of uniformly-continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology.
-
The reader knows a definition of continuous, topological spaces map.
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The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
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The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of \(1\), and any positive natural number, there is a symmetric neighborhood of \(1\) whose power to the natural number is contained in the neighborhood.
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The reader admits the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element.
-
The reader admits the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point.
-
The reader admits the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Target Context
-
The reader will have a description and a proof of the proposition that for any continuous map from any group with topology with continuous operations (especially, topological group) into any normed vectors space with the induced topology, if the preimage of the nonzeros is contained in any compact subset of the domain, the map is uniformly-continuous.
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:
\(G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\(V\): \(\in \{\text{ the normed vectors spaces }\}\) with the induced topology induced by the metric induced by the norm
\(f\): \(: G \to V\), \(\in \{\text{ the continuous maps }\}\)
//
Statements:
\(\exists K \in \{\text{ the compact subsets of } G\} (f^{-1} (V \setminus \{0\}) \subseteq K)\)
\(\implies\)
\(f \in \{\text{ the uniformly-continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(k \in K\), take an open neighborhood of \(k\), \(U_k\), such that \(f (U_k) \subseteq B_{f (k), \epsilon / 2}\); Step 2: take a symmetric neighborhood of \(1\), \(W_{1, k}\), such that \(W_{1, k} W_{1, k} k \subseteq U_k\); Step 3: take a finite cover of \(K\), \(\{W_{1, k_j} k_j \vert j \in J\}\); Step 4: take \(W_1 := \cap_{j \in J} W_{1, k_j}\); Step 5: see that for each \(g, g' \in G\) such that \(g g'^{-1} \in W_1\), \(\Vert f (g) - f (g') \Vert \lt \epsilon\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(k \in K\) be any.
There is an open neighborhood of \(k\), \(U_k \subseteq G\), such that \(f (U_k) \subseteq B_{f (k), \epsilon / 2}\), because \(f\) is continuous.
Step 2:
There is a symmetric neighborhood of \(1\), \(W'_{1, k}\), such that \(W'_{1, k} k {W'_{1, k}}^{-1} \subseteq U_k\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element, then, \(W'_{1, k} k \subseteq W'_{1, k} k {W'_{1, k}}^{-1} \subseteq U_k\).
There is a symmetric neighborhood of \(1\), \(W_{1, k}\), such that \(W_{1, k} W_{1, k} \subseteq W'_{1, k}\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of \(1\), and any positive natural number, there is a symmetric neighborhood of \(1\) whose power to the natural number is contained in the neighborhood.
So, \(W_{1, k} W_{1, k} k \subseteq U_k\), so, \(W_{1, k} W_{1, k} \subseteq U_k k^{-1}\).
Also, \(W_{1, k} k \subseteq U_k\), because \(W_{1, k} \subseteq W_{1, k} W_{1, k}\), because for each \(w \in W_{1, k}\), \(w = w 1 \in W_{1, k} W_{1, k}\), so, \(W_{1, k} k \subseteq W_{1, k} W_{1, k} k\).
Step 3:
Each \(W_{1, k} k \subseteq G\) is a neighborhood of \(k\), by the proposition that for any group with any topology with any continuous operations (especially, any topological group), any element multiplied by any (open) neighborhood of \(1\) from left or right is a (open) neighborhood of the element, so, there is an open neighborhood of \(k\), \(V_k \subseteq G\), such that \(k \in V_k \subseteq W_{1, k} k\).
\(\{V_k \vert k \in K\}\) covers \(K\), so, \(\{V_k \vert k \in K\}\) is an open cover of \(K\).
As \(K\) is compact, there is a finite subcover, \(\{V_{k_j} \vert j \in J\}\).
\(\{W_{1, k_j} k_j \vert j \in J\}\) is a finite cover of \(K\).
Step 4:
Let us take \(W_1 := \cap_{j \in J} W_{1, k_j} \subseteq G\), which is a symmetric neighborhood of \(1\), by the proposition that for any topological space and its any point, the intersection of any finite number of neighborhoods of the point is a neighborhood of the point and the proposition that for any group, the intersection of any symmetric subsets is symmetric.
Step 5:
Let \(g, g' \in G\) be any such that \(g g'^{-1} \in W_1\).
When \(g, g' \notin K\), \(\Vert f (g) - f (g') \Vert = \Vert 0 - 0 \Vert = 0 \lt \epsilon\).
Let us suppose that \(g' \in K\).
\(g' \in W_{1, k_j} k_j \subseteq U_{k_j}\) for a \(j \in J\).
So, \(g' {k_j}^{-1} \in W_{1, k_j}\).
\(g g'^{-1} \in W_{1, k_j}\), because \(g g'^{-1} \in W_1 \subseteq W_{1, k_j}\).
\(g {k_j}^{-1} = (g g'^{-1}) (g' {k_j}^{-1}) \in W_{1, k_j} W_{1, k_j} \subseteq U_{k_j} {k_j}^{-1}\).
So, \(g \in U_{k_j}\).
As \(g, g' \in U_{k_j}\), \(\Vert f (g) - f (g') \Vert = \Vert f (g) - f (k_j) + f (k_j) - f (g') \Vert \le \Vert f (g) - f (k_j) \Vert + \Vert f (k_j) - f (g') \Vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
Let us suppose that \(g \in K\).
\(g \in W_{1, k_j} k_j \subseteq U_{k_j}\) for a \(j \in J\).
So, \(g {k_j}^{-1} \in W_{1, k_j}\).
\(g' g^{-1} \in W_{1, k_j}\), because \(g' g^{-1} = (g g'^{-1})^{-1} \in {W_1}^{-1} \subseteq W_{1, k_j}^{-1} = W_{1, k_j}\).
\(g' {k_j}^{-1} = (g' g^{-1}) (g {k_j}^{-1}) \in W_{1, k_j} W_{1, k_j} \subseteq U_{k_j} {k_j}^{-1}\).
So, \(g' \in U_{k_j}\).
As \(g, g' \in U_{k_j}\), \(\Vert f (g) - f (g') \Vert = \Vert f (g) - f (k_j) + f (k_j) - f (g') \Vert \le \Vert f (g) - f (k_j) \Vert + \Vert f (k_j) - f (g') \Vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
So, anyway, \(\Vert f (g) - f (g') \Vert \lt \epsilon\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of uniformly continuous map from group with topology with continuous operations (especially, topological group) into normed vectors space with induced topology
The table of contents of this article
Main Body
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and subset of subspace, closure of subset on base space contained in subspace is closure on subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, the closure of the subset on the base space contained in the subspace is the closure of the subset on the subspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(S\): \(\subseteq T\)
\(\overline{S}^{T'}\): \(= \text{ the closure of } S \text{ on } T'\) such that \(\overline{S}^{T'} \subseteq T\)
\(\overline{S}^T\): \(= \text{ the closure of } S \text{ on } T\)
//
Statements:
\(\overline{S}^{T'} = \overline{S}^T\)
//
2: Proof
Whole Strategy: Step 1: apply the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace.
Step 1:
By the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace, \(\overline{S}^T = \overline{S}^{T'} \cap T = \overline{S}^{T'}\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
