$A$ be a $2\times 2$ real matrix with trace $2$ and determinant $-3$

$A$ be a $2\times 2$ real matrix with trace $2$ and determinant $-3$, consider the linear map $T:M_2(\mathbb{R})\to M_2(\mathbb{R}):=B\to AB$ Then which of the following are true?

1. $T$ is diagonalizable

2. $T$ is invertible

3. $2$ is an eigen value of $T$

4. $T(B)=B$ for some $0\ne B\in M_2(\mathbb{R})$

if $ A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$ Then I have calculated that matrix of $T$ will be \begin{pmatrix}a_{11}&a_{12}&0&0\\a_{21}&a_{22}&0&0\\0&0&a_{11}&a_{12}\\0&0&a_{21}&a_{22}\end{pmatrix} so $T$ is invertible I can say as $\det T=(\det A)^2=9$, could any one help to find out others as true/false?

Solution 1:

Nicely done so far. Don't forget to precise that you took the matrix of $T$ with respect to the basis $\{E_{11},E_{12},E_{21},E_{22}\}$ if you are to write this down.

By assumptions, the characteristic polynomial of $A$ is $$X^2-(\mbox{tr} \,A)X+\det A=X^2-2X-3=(X-3)(X+1).$$ So the eigenvalues of $A$, $-1,3$, are distinct. Hence $A$ is diagonalizable. This should help you answer 1, 3, and 4. By block considerations. Note that 4 says that $1$ is an eigenvalue of $T$.

In case you are not used to matrix block computations, if $P$ is invertible in $M_2$, then $$ \pmatrix{P&0\\0&P}\pmatrix{A&0\\0&A}\pmatrix{P^{-1}&0\\0&P^{-1}}=\pmatrix{PAP^{-1}&0\\0&PAP^{-1}} $$ in $M_4$. So if $P$ diagonalizes $A$...

Solution 2:

claim:

minimal polynomial of $T$ and the minimal polynomial of $A$ are the same .

let $p(x)=x^l+s_{l-1}x^{l-1}\ldots+ s_0$ (minimal polynomial of $A$) .

$p(A)=0\implies p(A)B=0 \implies A^lB+s_{l-1}A^{l-1}B+\ldots+s_0B=0$

which gives $$ T^lB+s_{l-1}T^{l-1}B+\ldots+s_0B=0$$

hence minimal polynomial of $T$ $ |$ minimal polynomial of $A$

similarly it can be shown minimal polynomial of $A$ $|$ minimal polynomial of $T$.

so as $A$ is diagonalizable $\implies $ $T$ is diagonalizable

$2,1$ are also not eigenvalues of $A$ and hence they are not eigenvalues of $T$ so option $3,4 $ are false

Solution 3:

T(B)=0 i.e AB=0,but A is invertible.This implies B=0. Hence kerT={0}. T is one-one.T is Onto also.Hence T is invertible.