Showing posts with label Definitions and Propositions. Show all posts
Showing posts with label Definitions and Propositions. Show all posts

2026-09-21

2005: Conjugate of Hermitian Matrix by Unitary Matrix Is Hermitian

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

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 the conjugate of any Hermitian matrix by any unitary matrix is Hermitian.

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:
\(M\): \(\in \{\text{ the Hermitian matrices }\}\)
\(N\): \(\in \{\text{ the unitary matrices }\}\)
//

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

2: Proof


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

Step 1:

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

So, \(N M N^{- 1}\) is valid.

Step 2:

We hereafter use the proposition that for any ring, the multiplications of any matrices over the ring are associative.

\((N M N^{- 1})^* = {N^{- 1}}^* M^* N^*\), by the proposition that the Hermitian conjugate of the product of any complex matrices is the product of the Hermitian conjugates of the constituents in the reverse order, \(= N M N^{- 1}\), by the proposition that for any unitary matrix, the Hermitian conjugate of the inverse of the matrix and the inverse of the Hermitian conjugate of the matrix is the matrix.

So, \((N M N^{- 1})^*\) is Hermitian.

References
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2004: Conjugate of Unitary Matrix by Unitary Matrix Is Unitary

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

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 the conjugate of any unitary matrix by 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:
\(M\): \(\in \{\text{ the unitary matrices }\}\)
\(N\): \(\in \{\text{ the unitary matrices }\}\)
//

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

2: Proof


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

Step 1:

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

So, \(N M N^{- 1}\) is valid.

Step 2:

We hereafter use the proposition that for any ring, the multiplications of any matrices over the ring are associative.

\((N M N^{- 1})^* = {N^{- 1}}^* M^* N^*\), by the proposition that the Hermitian conjugate of the product of any complex matrices is the product of the Hermitian conjugates of the constituents in the reverse order.

So, \((N M N^{- 1})^* N M N^{- 1} = {N^{- 1}}^* M^* N^* N M N^{- 1} = {N^{- 1}}^* M^* (N^* N) M N^{- 1} = {N^{- 1}}^* M^* I M N^{- 1} = {N^{- 1}}^* (M^* M) N^{- 1} = {N^{- 1}}^* I N^{- 1} = {N^{- 1}}^* N^{- 1} = I\), by the proposition that the inverse of any unitary matrix is unitary.

Likewise, \(N M N^{- 1} (N M N^{- 1})^* = N M N^{- 1} {N^{- 1}}^* M^* N^* = N M (N^{- 1} {N^{- 1}}^*) M^* N^* = N M I M^* N^*\), by the proposition that the inverse of any unitary matrix is unitary, \(= N (M M^*) N^* = N I N^* = N N^* = I\).

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

So, \((N M N^{- 1})^*\) is unitary.

References
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2003: For Unitary Matrix, Hermitian Conjugate of Inverse of Matrix and Inverse of Hermitian Conjugate of Matrix Is Matrix

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description/proof of that for unitary matrix, Hermitian conjugate of inverse of matrix and inverse of Hermitian conjugate of matrix is 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 for any unitary matrix, the Hermitian conjugate of the inverse of the matrix and the inverse of the Hermitian conjugate of the 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:
\(M\): \(\in \{\text{ the unitary matrices }\}\)
//

Statements:
\({M^{- 1}}^* = {M^*}^{- 1} = M\)
//

2: Proof


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

Step 1:

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

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

References
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2002: Conjugate of Matrix by Invertible Matrix

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definition of conjugate of matrix by invertible matrix

Topics


About: ring
About: matrices space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of conjugate of matrix by invertible 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:
\( R\): \(\in \{\text{ the rings }\}\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( M_n (R)\): \(= \text{ the ring of the } n \times n R \text{ matrices }\)
\( M\): \(\in M_n (R)\)
\( N\): \(\in \{\text{ the invertible matrices in } M_n (R)\}\)
\(*N M N^{- 1}\):
//

Conditions:
//

2: Note


A main reason why we take up this concept is that it appears in the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this.

References
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2001: For Module with \(2\) Bases of Same Finite Cardinality and Module Endomorphism, Transition of Endomorphism Matrices w.r.t. Change of Bases Is This

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description/proof of that for module with \(2\) bases of same finite cardinality and module endomorphism, transition of endomorphism matrices w.r.t. change of bases is this

Topics


About: module

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 module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this.

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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(B\): \(\in \{\text{ the bases for } M\} = \{b^j \vert j \in \{1, ..., n\}\}\)
\(B'\): \(\in \{\text{ the bases for } M\} = \{b'^l = N^l_j b^j \vert l \in \{1, ..., n\}\}\)
\(f\): \(: M \to M\), \(\in \{\text{ the ring endomorphisms }\}\)
\(O\): \(= \text{ the matrix of } f \text{ with respect to } B\) defined in 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 \(d_1 \times d_2\) ring matrix
\(O'\): \(= \text{ the matrix of } f \text{ with respect to } B'\) defined in 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 \(d_1 \times d_2\) ring matrix
//

Statements:
\(O' = N O N^{-1}\)
//

2: Note


This holds not only for any vectors space \(M\) but also for any module \(M\) as far as such bases exist.

3: Proof


Whole Strategy: Step 1: see that \(N\) is uniquely determined; Step 2: take \(N' \in M_n (R)\) such that \(b^j = N'^j_l b'^l\), and see that \(N' = N^{- 1}\); Step 3: see that \(f (b'^l) = N^l_j O^j_m N'^m_p b'^p\).

Step 1:

\(N\) 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.

Step 2:

There is the unique \(N' \in M_n (R)\) such that \(b^j = N'^j_l b'^l\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

For each \(l \in \{1, ..., n\}\), \(b'^l = N^l_j b^j = N^l_j N'^j_m b'^m\), which implies that \(N^l_j N'^j_l = 1\) and \(N^l_j N'^j_m = 0\) for each \(m \neq l\), which implies that \(N N' = I\).

For each \(j \in \{1, ..., n\}\), \(b^j = N'^j_l b'^l = N'^j_l N^l_m b^m\), which implies that \(\) and \(N'^j_l N^l_j = 1\) and \(N'^j_l N^l_m = 0\) for each \(m \neq j\), which implies that \(N' N = I\).

So, \(N'\) is an inverse of \(N\) and is the inverse, \(N^{- 1}\), by the proposition that if any square ring matrix has an inverse, the inverse is the unique inverse.

Step 3:

For each \(l \in \{1, ..., n\}\), \(f (b'^l) = f (N^l_j b^j) = N^l_j f (b^j)\), because \(f\) is linear, \(= N^l_j O^j_m b^m = N^l_j O^j_m N'^m_p b'^p = O'^l_p b'^p\), which implies that \(N^l_j O^j_m N'^m_p = O'^l_p\), which implies that \(N O N' = O'\).

So, \(O' = N O N^{- 1}\).

References
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2000: If Square Ring Matrix Has Inverse, Inverse Is Unique

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description/proof of that if square ring matrix has inverse, inverse is unique

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 if any square ring matrix has an inverse, the inverse is the unique inverse.

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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_n (R)\): \(= \text{ the ring of the } n \times n R \text{ matrices }\)
\(M\): \(\in M_n (R)\)
//

Statements:
\(\exists M' \in M_n (R) (M' M = M M' = I) \land \exists M'' \in M_n (R) (M'' M = M M'' = I)\)
\(\implies\)
\(M' = M''\)
//

2: Proof


Whole Strategy: Step 1: apply the proposition that for any ring, if an element has an inverse, the inverse is unique.

Step 1:

As \(M_n (R)\) is a ring, by Note for the definition of ring of \(n \times n\) ring matrices, the proposition holds, by the proposition that for any ring, if an element has an inverse, the inverse is unique.

References
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1999: Component of Unitary Matrix Has Absolute Value Equal to or Smaller than \(1\)

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description/proof of that component of unitary matrix has absolute value equal to or smaller than \(1\)

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 each component of any unitary matrix has an absolute value equal to or smaller than \(1\).

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:
\(M\): \(\in \{\text{ the } n \times n \text{ unitary matrices }\}\)
//

Statements:
\(\forall j, l \in \{1, ..., n\} (\vert M^j_l \vert \le 1)\)
//

2: Proof


Whole Strategy: Step 1: see that \(\sum_{l \in \{1, ..., n\}} \vert M^j_l \vert^2 = 1\).

Step 1:

\(M M^* = I\), by the definition of unitary matrix.

Let \(j \in \{1, ..., n\}\) be any.

\((M M^*)^j_j = I^j_j = 1\), but he left hand side is \(\sum_{l \in \{1, ..., n\}} M^j_l {M^*}^l_j = \sum_{l \in \{1, ..., n\}} M^j_l \overline{M^j_l} = \sum_{l \in \{1, ..., n\}} \vert M^j_l \vert^2\).

So, for each \(l \in \{1, ..., n\}\), \(\vert M^j_l \vert^2 \le 1\).

So, \(\vert M^j_l \vert \le 1\) for each \(j, l \in \{1, ..., n\}\).

References
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1998: \(n \times n\) Unitary Matrices Group

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definition of \(n \times n\) unitary matrices group

Topics


About: group
About: matrices space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of \(n \times n\) unitary matrices 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:
\( \mathbb{C}\): \(= \text{ the complex numbers field }\)
\( \{M\}\): \(= \text{ the } \mathbb{C} \text{ matrices space }\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\(*U (n)\): \(= \{M \in \{M\} \vert M \in \{\text{ the } n \times n \text{ unitary matrices }\}\}\)
//

Conditions:
//

2: Note


Let us see that \(U (n)\) is indeed a group.

Let \(M_1, M_2, M_3 \in U (n)\) be any.

\(M_1 M_2 \in U (n)\), because \((M_1 M_2)^* = {M_2}^* {M_1}^*\), by the proposition that the Hermitian conjugate of the product of any complex matrices is the product of the Hermitian conjugates of the constituents in the reverse order, \(= {M_2}^{- 1} {M_1}^{- 1}\), so, \((M_1 M_2)^* M_1 M_2 = {M_2}^{- 1} {M_1}^{- 1} M_1 M_2 = {M_2}^{- 1} I M_2 = {M_2}^{- 1} M_2 = I\) and \(M_1 M_2 (M_1 M_2)^* = M_1 M_2 {M_2}^{- 1} {M_1}^{- 1} = M_1 I {M_1}^{- 1} = M_1 {M_1}^{- 1} = I\), so, \((M_1 M_2)^* = (M_1 M_2)^{- 1}\).

1) \((M_1 \bullet M_2) \bullet M_3 = M_1 \bullet (M_2 \bullet M_3)\): by the proposition that for any ring, the multiplications of any matrices over the ring are associative.

2) \(i \in U (n)\) (called 'identity element') such that \(i \bullet M_1 = M_1 \bullet i = M_1\): the identity matrix, \(I\), is in \(U (n)\), because \(I^* = I = I^{-1}\), and \(I M_1 = M_1 I = M_1\).

3) \({M_1}^{- 1} \in U (n)\) (called 'inverse element of \(M_1\)') such that \({M_1}^{- 1} \bullet M_1 = M_1 \bullet {M_1}^{- 1} = I\): \({M_1}^* = {M_1}^{- 1} \in U (n)\), because \({{M_1}^*}^* = M_1\) and \({M_1}^* M_1 = I = M_1 {M_1}^*\), so, \({{M_1}^*}^{- 1} = M_1 = {{M_1}^*}^*\).

So, \(U (n)\) is a group.

References
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1997: \(n \times n\) Orthogonal Matrices Group

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definition of \(n \times n\) orthogonal matrices group

Topics


About: group
About: matrices space

The table of contents of this article

Starting Context

Target Context


  • The reader will have a definition of \(n \times n\) orthogonal matrices 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:
\( \mathbb{R}\): \(= \text{ the real numbers field }\)
\( \{M\}\): \(= \text{ the } \mathbb{R} \text{ matrices space }\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\(*O (n)\): \(= \{M \in \{M\} \vert M \in \{\text{ the } n \times n \text{ orthogonal matrices }\}\}\)
//

Conditions:
//

2: Note


Let us see that \(O (n)\) is indeed a group.

Let \(M_1, M_2, M_3 \in O (n)\) be any.

\(M_1 M_2 \in O (n)\), because \((M_1 M_2)^t = {M_2}^t {M_1}^t\), 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, \(= {M_2}^{- 1} {M_1}^{- 1}\), so, \((M_1 M_2)^t M_1 M_2 = {M_2}^{- 1} {M_1}^{- 1} M_1 M_2 = {M_2}^{- 1} I M_2 = {M_2}^{- 1} M_2 = I\) and \(M_1 M_2 (M_1 M_2)^t = M_1 M_2 {M_2}^{- 1} {M_1}^{- 1} = M_1 I {M_1}^{- 1} = M_1 {M_1}^{- 1} = I\), so, \((M_1 M_2)^t = (M_1 M_2)^{- 1}\).

1) \((M_1 \bullet M_2) \bullet M_3 = M_1 \bullet (M_2 \bullet M_3)\): by the proposition that for any ring, the multiplications of any matrices over the ring are associative.

2) \(i \in O (n)\) (called 'identity element') such that \(i \bullet M_1 = M_1 \bullet i = M_1\): the identity matrix, \(I\), is in \(O (n)\), because \(I^t = I = I^{-1}\), and \(I M_1 = M_1 I = M_1\).

3) \({M_1}^{- 1} \in O (n)\) (called 'inverse element of \(M_1\)') such that \({M_1}^{- 1} \bullet M_1 = M_1 \bullet {M_1}^{- 1} = I\): \({M_1}^t = {M_1}^{- 1} \in O (n)\), because \({{M_1}^t}^t = M_1\) and \({M_1}^t M_1 = I = M_1 {M_1}^t\), so, \({{M_1}^t}^{- 1} = M_1 = {{M_1}^t}^t\).

So, \(O (n)\) is a group.

References
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1996: Module Has Unique \(0\) Element

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description/proof of that module has unique \(0\) element

Topics


About: module

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 module has the unique \(0\) element.

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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//

Statements:
\(\forall m \in \{\text{ the zero elements of } M\} (m = 0)\)
//

2: Proof


Whole Strategy: Step 1: see that \(m = 0\).

Step 1:

The definition of module requires the existence of a \(0 \in M\), but does not directly require that it is the unique zero element.

Let \(m \in M\) be any zero element.

\(m + 0 = m\), because \(0\) is a zero element, but \(m + 0 = 0\), because \(m\) is a zero element.

So, \(m = 0\).

References
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1995: For Linear Map Between Modules with Same Cardinality Bases, Map Is Bijection iff Corresponding Matrix Is Invertible

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description/proof of that for linear map between modules with same cardinality bases, map is bijection iff corresponding matrix is invertible

Topics


About: module

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 linear map between any modules with any same cardinality bases, the map is a bijection if and only if the corresponding matrix with respect to the bases is invertible.

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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules } \}\), with any basis, \(B_1 = \{{b_1}^1, ..., {b_1}^d\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules } \}\), with any basis, \(B_2 = \{{b_2}^1, ..., {b_2}^d\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
\(M\): \(\in M_n (R)\), \(= \text{ the matrix }\) which corresponds to \(f\) with respect to \(B_1\) and \(B_2\) defined in 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 \(d_1 \times d_2\) ring matrix
//

Statements:
\(f \in \{\text{ the bijections }\}\)
\(\iff\)
\(M \in \{\text{ the invertible matrices }\}\)
//

2: Proof


Whole Strategy: Step 0: see what \(M\) is; Step 1: suppose that \(f\) is a bijection; Step 2: see that \(M\) is invertible; Step 3: suppose that \(M\) is invertible; Step 4: see that \(f\) is a bijection.

Step 0:

As \(f\) is linear, \(f\) is represented by \(M\), 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 \(d_1 \times d_2\) ring matrix.

\(M\) is determined by that \(f ({b_1}^j) = M^j_l {b_2}^l\).

Step 1:

Let us suppose that \(f\) is a bijection.

Step 2:

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

So, there is the linear inverse, \(f^{- 1}: M_2 \to M_1\).

As \(f^{- 1}\) is linear, \(f^{- 1}\) is represented by an \(M' \in M_n (R)\), 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 \(d_1 \times d_2\) ring matrix.

As \(f^{- 1} \circ f = id_{M_1}\) is linear, \(f^{- 1} \circ f\) is represented by an \(M'' \in M_n (R)\), 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 \(d_1 \times d_2\) ring matrix.

\(f^{- 1} \circ f (m_1) = f^{- 1} \circ f ({m_1}_j {b_1}^j) = {m_1}_j M''^j_l {b_1}^l\), but it equals \(id_{M_1} (m_1) = m_1 = {m_1}_l {b_1}^l\), which means that \({m_1}_j M''^j_l = {m_1}_l\), which means that \({m_1}^t M'' = {m_1}^t\), so, \(M'' = I\), by the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.

On the other hand, \(f^{- 1} \circ f (m_1) = f^{- 1} (f (m_1)) = f^{- 1} (f ({m_1}_j {b_1}^j)) = f^{- 1} ({m_1}_j M^j_l {b_2}^l) = {m_1}_j M^j_l M'^l_p {b_1}^p\), which means that \({m_1}_j M''^j_p = {m_1}_j M^j_l M'^l_p\), which implies that \(M''^j_p = M^j_l M'^l_p\) (take \({m_1}_j = 1\) and the other components \(0\)), which means that \(M'' = M M'\).

So, \(M M' = I\).

As \(f \circ f^{- 1} = id_{M_2}\) is linear, \(f \circ f^{- 1}\) is represented by an \(M'' \in M_n (R)\), 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 \(d_1 \times d_2\) ring matrix.

\(f \circ f^{- 1} (m_2) = f \circ f^{- 1} ({m_2}_j {b_2}^j) = {m_2}_j M''^j_l {b_2}^l\), but it equals \(id_{M_2} (m_2) = m_2 = {m_2}_l {b_2}^l\), which means that \({m_2}_j M''^j_l = {m_2}_l\), which means that \({m_2} M'' = {m_2}\), so, \(M'' = I\), by the proposition that for any \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.

On the other hand, \(f \circ f^{- 1} (m_2) = f (f^{- 1} (m_2)) = f (f^{- 1} ({m_2}_j {b_2}^j)) = f ({m_2}_j M'^j_l {b_1}^l) = {m_2}_j M'^j_l M^l_p {b_2}^p\), which means that \({m_2}_j M''^j_p = {m_2}_j M'^j_l M^l_p\), which implies that \(M''^j_p = M'^j_l M^l_p\) (take \({m_2}_j = 1\) and the other components \(0\)), which means that \(M'' = M' M\).

So, \(M' M = I\).

So, \(M\) is invertible with the inverse, \(M'\).

Step 3:

Let us suppose that \(M\) is invertible.

Step 4:

There is an \(M' \in M_n (R)\) such that \(M' M = I\) and \(M M' = I\).

\(f': M_2 \to M_1\) defined as \(f' ({m_2}_j {b_2}^j) = {m_2}_j M'^j_l {b_1}^l\) 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 \(d_1 \times d_2\) ring matrix.

\(f' \circ f (m_1) = f' (f (m_1)) = f' (f ({m_1}_j {b_1}^j)) = f' ({m_1}_j M^j_l {b_2}^l) = {m_1}_j M^j_l M'^l_p {b_1}^p = {m_1}_j (M M')^j_p {b_1}^p = {m_1}_j I^j_p {b_1}^p = {m_1}_j {b_1}^j = m_1\), which means that \(f' \circ f = id_{M_1}\).

\(f \circ f' (m_2) = f (f' (m_2)) = f (f' ({m_2}_j {b_2}^j)) = f ({m_2}_j M'^j_l {b_1}^l) = {m_2}_j M'^j_l M^l_p {b_2}^p = {m_2}_j (M' M)^j_p {b_2}^p = {m_2}_j I^j_p {b_2}^p = {m_2}_j {b_2}^j = m_2\), which means that \(f \circ f' = id_{M_2}\).

So, \(f'\) is an inverse of \(f\).

\(f\) is a bijection, by the proposition that any map is a bijection if and only if it has an inverse.

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1994: For \(n \times n\) Ring Matrix, iff Matrix Multiplied by Each Column or Row Is Original Column or Row, Matrix Is Identity

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description/proof of that for \(n \times n\) ring matrix, iff matrix multiplied by each column or row is original column or row, matrix is identity

Topics


About: ring

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 \(n \times n\) ring matrix, if and only if the matrix multiplied by each column or row is the original column or row, the matrix is the identity.

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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in M_n (R)\)
//

Statements:
(
\(\forall m \in R^m (M m = m)\), where \(m\) is regarded to be the column
\(\iff\)
\(M = I\)
)
\(\land\)
(
\(\forall m \in R^m (m M = m)\), where \(m\) is regarded to be the row
\(\iff\)
\(M = I\)
)
//

2: Proof


Whole Strategy: Step 1: suppose that \(\forall m \in R^m (M m = m)\); Step 2: see that \(M = I\); Step 3: suppose that \(M = I\); Step 4: see that \(\forall m \in R^m (M m = m)\); Step 5: suppose that \(\forall m \in R^m (m M = m)\); Step 6: see that \(M = I\); Step 7: suppose that \(M = I\); Step 8: see that \(\forall m \in R^m (m M = m)\).

Step 1:

Let us suppose that \(\forall m \in R^m (M m = m)\).

Step 2:

\(M^j_l m^l = m^j\).

Let \(p \in \{1, ..., n\}\) be any.

Let us take \(m\) such that \(m^p = 1\) and \(m^j = 0\) for each \(j \neq p\).

\(M^p_l m^l = m^p = 1\), but the left hand side is \(M^p_p m^p = M^p_p 1 = M^p_p\), so, \(M^p_p = 1\).

For each \(j \neq p\), \(M^j_l m^l = m^j = 0\), but the left hand side is \(M^j_p m^p = M^j_p 1 = M^j_p\), so, \(M^j_p = 0\).

So, each diagonal component of \(M\) is \(1\) and the other components of \(M\) are \(0\).

So, \(M = I\).

Step 3:

Let us suppose that \(M = I\).

Step 4:

For each \(m \in R^m\), \((M m)^j = M^j_l m^l = I^j_l m^l = m^j\).

So, \(M m = m\).

Step 5:

Let us suppose that \(\forall m \in R^m (m M = m)\).

Step 6:

\(m_l M^l_j = m_j\).

Let \(p \in \{1, ..., n\}\) be any.

Let us take \(m\) such that \(m_p = 1\) and \(m_j = 0\) for each \(j \neq p\).

\(m_l M^l_p = m_p = 1\), but the left hand side is \(m_p M^p_p = 1 M^p_p = M^p_p\), so, \(M^p_p = 1\).

For each \(j \neq p\), \(m_l M^l_j = m_j = 0\), but the left hand side is \(m_p M^p_j = 1 M^p_j = M^p_j\), so, \(M^p_j = 0\).

So, each diagonal component of \(M\) is \(1\) and the other components of \(M\) are \(0\).

So, \(M = I\).

Step 7:

Let us suppose that \(M = I\).

Step 8:

For each \(m \in R^m\), \((m M)_j = m_l M^l_j = m_l I^l_j = m_j\).

So, \(m M = m\).

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1993: Finite Product Module of Modules with Bases Has This Basis

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description/proof of that finite product module of modules with bases has this basis

Topics


About: module

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 finite product module of modules with bases has this 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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(R\): \(\in \{\text{ the rings }\}\)
\(\{M_j \vert j \in J\}\): \(\subseteq \{\text{ the } R \text{ modules }\}\)
\(\times_{j \in J} M_j\): \(= \text{ the product module }\)
//

Statements:
\(\forall j \in J (B_j = \{b_{j, l_j} \vert l_j \in L_j\} \in \{\text{ the bases for } M_j\})\) where \(L_j\) is a possibly uncountable index set
\(\implies\)
\(B := \cup_{j \in J} \{\times_{j' \in J} \delta_{j, j'} b_{j, l_j} \vert l_j \in L_j\} \in \{\text{ the bases for } \times_{j \in J} M_j \}\)
//

2: Note


\(B\) is like \(\{(b_{1, l_1}, 0, ..., 0) \vert l_1 \in L_1\} \cup ... \cup \{(0, ..., 0, b_{n, l_n}) \vert l_n \in L_n\}\).

\(J\) needs to be finite, because otherwise, an \(m \in \times_{j \in J} M_j\) would not be any finite linear combination of \(B\).

3: Proof


Whole Strategy: Step 1: see that \(B\) is linearly independent; Step 2: see that each \(m \in \times_{j \in J} M_j\) is a finite linear combination of \(B\); Step 3: conclude the proposition.

Step 1:

Let us see that \(B\) is linearly independent.

Let \(S \subseteq B\) be any finite subset.

\(S = \cup_{j \in J} \{\times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} \vert l^`_j \in L^`_j\}\), where \(L^`_j \subseteq L_j\) is a finite subset possibly empty: when \(L^`_j\) is empty, \(\{\times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} \vert l^`_j \in L^`_j\}\) is empty, which means that no element of \(M_j\) has not been chosen.

Let \(\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l^`_j} = 0\).

Then, for each \(j'' \in J\), \((\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l^`_j}) (j'') = 0\), but the left hand side is \(\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \delta_{j, j''} b_{j, l^`_j} = \sum_{l^`_{j''} \in L^`_{j''}} r_{{j''}, l^`_{j''}} b_{j'', l^`_{j''}}\), which implies that \(r_{{j''}, l^`_{j''}} = 0\) for each \(j'' \in J\) and each \(l^`_{j''} \in L^`_{j''}\), because \(B_{j''}\) is linearly independent.

That means that all the \(r_{j, l^`_j}\) s are \(0\).

So, \(B\) is linearly independent.

Step 2:

Let \(m \in \times_{j \in J} M_j\) be any.

For each \(j \in J\), \(m (j) = \sum_{l^`_j \in L^`_j} r_{j, l^`_j} b_{j, l^`_j}\), where \(L^`_j \subseteq L_j\) is a finite subset, because \(B_j\) is a basis for \(M_j\).

\(m = \sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l_j}\), because for each \(j'' \in J\), \((\sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \times_{j' \in J} \delta_{j, j'} b_{j, l_j}) (j'') = \sum_{j \in J, l^`_j \in L^`_j} r_{j, l^`_j} \delta_{j, j''} b_{j, l_j} = \sum_{l^`_{j''} \in L^`_{j''}} r_{j'', l^`_{j''}} b_{j'', l_{j''}} = m (j'')\).

So, \(m\) is a finite linear combination of \(B\).

Step 3:

So, \(B\) is a basis for \(\times_{j \in J} M_j\).

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1992: Field Is Canonically \(1\)-Dimensional Vectors Space

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description/proof of that field is canonically \(1\)-dimensional vectors space

Topics


About: field
About: vectors 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 field is canonically a \(1\)-dimensional vectors space.

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:
\(F\): \(\in \{\text{ the fields }\}\), with the addition, \(+: F \times F \to F, (r_1, r_2) \mapsto r_1 + r_2\), and the scalar multiplication, \(.: F \times F \to F, (r_1, r_2) \mapsto r_1 r_2\)
//

Statements:
\(F \in \{\text{ the } F \text{ vectors spaces }\}\)
\(\land\)
\(\{1\} \in \{\text{ the bases for } F\}\)
//

2: Proof


Whole Strategy: Step 1: see that \(F\) satisfies the conditions to be an \(F\) vectors space; Step 2: see that \(\{1\}\) is a basis for the vectors space, \(F\).

Step 1:

Let us see that \(F\) satisfies the conditions to be an \(F\) vectors space.

1) \(\forall r_1, r_2 \in F (r_1 + r_2 \in F)\) (closed-ness under addition): because \(F\) is a field, which is a ring, which is an Abelian group under addition.

2) \(\forall r_1, r_2 \in F (r_1 + r_2 = r_2 + r_1)\) (commutativity of addition): because \(F\) is a field, which is a ring, which is an Abelian group under addition.

3) \(\forall r_1, r_2, r_3 \in F ((r_1 + r_2) + r_3 = r_1 + (r_2 + r_3))\) (associativity of additions): because \(F\) is a field, which is a ring, which is an Abelian group under addition.

4) \(\exists 0 \in F (\forall r \in F (r + 0 = r))\) (existence of 0 element): because \(F\) is a field, which is a ring, which is an Abelian group under addition.

5) \(\forall r \in F (\exists r' \in F (r' + r = 0))\) (existence of inverse element): because \(F\) is a field, which is a ring, which is an Abelian group under addition.

6) \(\forall r \in F, \forall r' \in F (r . r' \in F)\) (closed-ness under scalar multiplication): because \(F\) is a field, which is a ring, which is a monoid under multiplication.

7) \(\forall r \in F, \forall r_1, r_2 \in F ((r_1 + r_2) . r = r_1 . r + r_2 . r)\) (scalar multiplication distributability for scalars addition): because \(F\) is a field, which is a ring.

8) \(\forall r_1, r_2 \in F, \forall r \in F (r . (r_1 + r_2) = r . r_1 + r . r_2)\) (scalar multiplication distributability for vectors addition): because \(F\) is a field, which is a ring.

9) \(\forall r \in F, \forall r_1, r_2 \in F ((r_1 r_2) . r = r_1 . (r_2 . r))\) (associativity of scalar multiplications): because \(F\) is a field, which is a ring, which is a monoid under multiplication.

10) \(\forall r \in F (1 . r = r)\) (identity of 1 multiplication): because \(F\) is a field, which is a ring, which is a monoid under multiplication.

So, \(F\) is an \(F\) vectors space.

Step 2:

\(\{1\} \subseteq F\) is linearly independent, because for \(r 1 = 0\), \(r 1 = r\), so, \(r = 0\).

For each \(r \in F\), \(r = r 1\).

So, \(\{1\}\) is a basis for the \(F\) vectors space, \(F\).

So, \(F\) is a \(1\)-dimensional vectors space.

References
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