given a linear transformation: $T:{M}_{n}(\mathbb{C})\to {M}_{n}(\mathbb{C})$, $T(A)=A-2{A}^{T}$, what is the

Joshua Foley
2022-07-16
Answered

representative matrix of a linear transformation

given a linear transformation: $T:{M}_{n}(\mathbb{C})\to {M}_{n}(\mathbb{C})$, $T(A)=A-2{A}^{T}$, what is the

given a linear transformation: $T:{M}_{n}(\mathbb{C})\to {M}_{n}(\mathbb{C})$, $T(A)=A-2{A}^{T}$, what is the

You can still ask an expert for help

Nicolas Calhoun

Answered 2022-07-17
Author has **15** answers

Let $\mathcal{A}$ denote the subspace of symmetric (antisymmetric matrices). Denote by ${E}_{ij}$ the matrix all of whose coefficients are zero except in the entry at the intersection of the i-th line and the $j$-th column, equal to 1. Thus ($({E}_{ij})$) is the canonical basis of $V$.

For any $i<j$ we have $\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}({E}_{ij},{E}_{ji})=\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}({S}_{ij},{A}_{ij})$ where ${S}_{ij}=\frac{{E}_{ij}+{E}_{ji}}{2}$ and ${A}_{ij}=\frac{{E}_{ij}-{E}_{ji}}{2}$

The transpose map $\tau :A\mapsto {A}^{T}$ is easily seen to be diagonal in that basis : we have

$\tau {E}_{ii}={E}_{ii},\tau {S}_{ij}={S}_{ij},\tau {A}_{ij}=-{A}_{ij}$

Note that $\mathcal{S}$ is the eigenspace of $\tau $ corresponding to the eigenvalue $1$. So

$\{{E}_{ii}|1\le i\le n\}\cup \{{S}_{ij}|1\le i<j\le n\}$ is a basis of $\mathcal{S}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S}\mathcal{)}\mathcal{=}\frac{\mathcal{n}\mathcal{(}\mathcal{n}\mathcal{+}\mathcal{1}\mathcal{)}}{\mathcal{2}}$<brSimilarly $\{{S}_{ij}|1\le i<j\le n\}$ is a basis of $\mathcal{A}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A}\mathcal{)}\mathcal{=}\frac{\mathcal{n}\mathcal{(}\mathcal{n}\mathcal{-}\mathcal{1}\mathcal{)}}{\mathcal{2}}$

For $A\in \mathcal{S}$ we have ${A}^{T}=A$ so $\sigma (A)=-A$.

For $A\in \mathcal{A}$ we have ${A}^{T}=-A$ so $\sigma (A)=5A$

So the characteristic polynomial ${\chi}_{A}$ of $A$ is

${\chi}_{A}=(X+1{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S})}(X-5{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A})}=(X+1{)}^{\frac{n(n+1)}{2}}(X-5{)}^{\frac{n(n+1)}{2}}$${\chi}_{A}=(X+1{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S})}(X-5{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A})}=(X+1{)}^{\frac{n(n+1)}{2}}(X-5{)}^{\frac{n(n+1)}{2}}$

For any $i<j$ we have $\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}({E}_{ij},{E}_{ji})=\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}({S}_{ij},{A}_{ij})$ where ${S}_{ij}=\frac{{E}_{ij}+{E}_{ji}}{2}$ and ${A}_{ij}=\frac{{E}_{ij}-{E}_{ji}}{2}$

The transpose map $\tau :A\mapsto {A}^{T}$ is easily seen to be diagonal in that basis : we have

$\tau {E}_{ii}={E}_{ii},\tau {S}_{ij}={S}_{ij},\tau {A}_{ij}=-{A}_{ij}$

Note that $\mathcal{S}$ is the eigenspace of $\tau $ corresponding to the eigenvalue $1$. So

$\{{E}_{ii}|1\le i\le n\}\cup \{{S}_{ij}|1\le i<j\le n\}$ is a basis of $\mathcal{S}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S}\mathcal{)}\mathcal{=}\frac{\mathcal{n}\mathcal{(}\mathcal{n}\mathcal{+}\mathcal{1}\mathcal{)}}{\mathcal{2}}$<brSimilarly $\{{S}_{ij}|1\le i<j\le n\}$ is a basis of $\mathcal{A}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A}\mathcal{)}\mathcal{=}\frac{\mathcal{n}\mathcal{(}\mathcal{n}\mathcal{-}\mathcal{1}\mathcal{)}}{\mathcal{2}}$

For $A\in \mathcal{S}$ we have ${A}^{T}=A$ so $\sigma (A)=-A$.

For $A\in \mathcal{A}$ we have ${A}^{T}=-A$ so $\sigma (A)=5A$

So the characteristic polynomial ${\chi}_{A}$ of $A$ is

${\chi}_{A}=(X+1{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S})}(X-5{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A})}=(X+1{)}^{\frac{n(n+1)}{2}}(X-5{)}^{\frac{n(n+1)}{2}}$${\chi}_{A}=(X+1{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{S})}(X-5{)}^{\mathsf{d}\mathsf{i}\mathsf{m}(\mathcal{A})}=(X+1{)}^{\frac{n(n+1)}{2}}(X-5{)}^{\frac{n(n+1)}{2}}$

asked 2021-09-13

Assume that A is row equivalent to B. Find bases for Nul A and Col A.

asked 2021-09-18

Find an explicit description of Nul A by listing vectors that span the null space.

asked 2021-06-13

For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.

$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$

Find a nonzero vector in Nul A.

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Find a nonzero vector in Nul A.

asked 2022-06-24

Let ${T}_{1}$ be a reflection of ${\mathbb{R}}^{3}$ in the xy plane, ${T}_{2}$ is a reflection of ${\mathbb{R}}^{3}$ in the xz plane. What is the standard matrix of transformation ${T}_{2}{T}_{1}$?

Here's my thinking so far:

Since the standard matrix for reflections in xy is

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

Similarly, standard matrix for orthogonal projection in the xz plane is

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$

I could multiply

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]\ast \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

to yield

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Could someone confirm for me if this is a valid approach?

Here's my thinking so far:

Since the standard matrix for reflections in xy is

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

Similarly, standard matrix for orthogonal projection in the xz plane is

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]$

I could multiply

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]\ast \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

to yield

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Could someone confirm for me if this is a valid approach?

asked 2022-01-30

Why is the Jacobian matrix equal to the matrix associated to a linear transformation?

Given the linear transformation f, we can construct the matrix A as follows: on the i-th column we put the vector$f\left({e}_{i}\right)$ where $E=({e}_{1},\cdots ,{e}_{n})$ is a basis of $\mathbb{R}}^{n$ .

Given the linear transformation f, we can construct the matrix A as follows: on the i-th column we put the vector

asked 2021-12-16

Is division of matrices possible?

Is it possible to divide a matrix by another? If yes, What will be the result of

asked 2021-09-18

Find $|u\cdot v|$ and determine whether $u\cdot v$

is directed into the screen or out of the screen.

$|u\cdot v|=??$

is directed into the screen or out of the screen.