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Let $A \in M_2(\mathbb{R})$ be of trace $2$ and determinant $-3$. Consider the linear transformation $T: M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$ by $$T(B)=AB$$ Then

1) $T$ is diagonalizable

2) $2$ is an eigenvalue of $T$

3) $T$ is invertible

4) $T(B)=B$, for some non zero $B$ in $M_2(\mathbb{R})$

My try:

3): $Ker(T)=\{B : T(B)=0\}=\{B : AB=0\}=\{0\}$, since $A$ is invertible and $AB=0 \Rightarrow B=0.$ Hence $T$ is one-one and so invertible. Therefore, 3) is true

4): Since there are so many matrices having trace $2$ and det $-3$, so i take $A$ as $A=\begin{bmatrix}1&2\\2&1\\\end{bmatrix}$ and
take $B=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}$ and so $T(B)=B$ implies $AB=B$. That is,

$$\begin{bmatrix}1&2\\2&1\\\end{bmatrix} \begin{bmatrix}a&b\\c&d\\\end{bmatrix}=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}$$

which implies $$a+2c=a$$ $$b+2d=b$$

$$2a+c=c$$

$$2b+d=d$$

Hence $a=b=c=d=0$ and so $B=0$. Therefore, 4) is false

2): Since the matrix of $T$ in the standard basis is

$$[T]=\begin{bmatrix}1&0&2&0\\0&1&0&2\\2&0&1&0\\0&2&0&1\end{bmatrix}$$ The eigenvalues of the above matrices are $-1$ and $3$, each with multiplicity $2$. Form this, 2) is false

1): $$\text{multiplicity of 3}=2=4-rank([T]-3I)$$ which is true

Similarly for the eigenvalue $-1$. so 1) is true

My Question is:

Is my all answers valid ?

Is there any short cut method to find each one ?

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5 Answers 5

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You solved 2 for a particular matrix. In general, the matrix of $T$ in the standard basis is\begin{bmatrix}\alpha&0&\beta&0\\0&\alpha&0&\beta\\\gamma&0&\delta&0\\0&\gamma&0&\delta\end{bmatrix}if $A=\left[\begin{smallmatrix}\alpha&\beta\\\gamma&\delta\end{smallmatrix}\right]$. So, the characteristic polynomial of $T$ is\begin{multline*}x^4-2 \alpha x^3-2 \delta x^3+\alpha^2 x^2+\delta^2 x^2-2 \beta \gamma x^2+4 \alpha \delta x^2-2 \alpha \delta^2 x+2 \alpha \beta \gamma x-2 \alpha^2 \delta x+\\+2 \beta \gamma \delta x+\beta^2 \gamma^2+\alpha^2 \delta^2-2 \alpha \beta \gamma \delta =\bigl((x-\alpha )(x-\delta )-\beta \gamma \bigr)^2=\\=\bigl(x^2-(\alpha +\delta )x+\alpha \delta -\beta \gamma \bigr)^2=\\=(x^2-\operatorname{tr}(A)x+\det(A))^2=(x^2-2x-3)^2\text,\end{multline*}whose roots are $-1$ and $3$.

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  • $\begingroup$ The solution of $2$ is actually perfectly fine. The statement is not true for one particular $A$, so it is not true for a general $A$. $\endgroup$
    – 5xum
    Commented Jul 11, 2017 at 10:00
  • $\begingroup$ @5xum You are right. I wrote what I wrote without noticing that. I shall not delete my answer though, because I think that it is interesting to notice that the answer is always the same, no matter which matrix $A$ you choose. $\endgroup$ Commented Jul 11, 2017 at 10:03
  • $\begingroup$ What if $M_2(\mathbb R)$ replaced by $M_n(\mathbb R)$? Will the characteristic polynomial of $T$ will be square of the characteristic polynomial of $A$? $\endgroup$ Commented Nov 24, 2019 at 2:52
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    $\begingroup$ No. It will be the characteristic polynomial of $A$ raised to the power $n$. $\endgroup$ Commented Nov 24, 2019 at 8:02
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Here's the short method: first, deduce that $A$ is diagonalizable since it has the distinct eigenvalues $-1,3$. Then, note that for any eigenvector $v$ of $A$ and any vector $x \in \Bbb R^2$, the matrix $vx^T \in M_2(\Bbb R)$ is an eigenvector of $T$. If we take $v_1,v_2$ to be the (linearly independent) eigenvectors of $A$, then the matrices $\{v_1v_1^T, v_1v_2^T, v_2v_1^T,v_2v_2^T\}$ form a basis of $M_2(\Bbb R)$ which diagonalizes $T$.

From there, the rest is easy.


Another perspective: let $T_A$ be the linear transformation $T_A(B) = AB$. If $A$ is diagonalizable with $A = SDS^{-1}$, then we have $$ T_A(B) = SDS^{-1}(B) = T_S(T_D(T_{S^{-1}}(B))) = [T_S \circ T_D \circ T_{S^{-1}}](B) $$ And verify that $T_D$ is diagonal with respect to "the standard basis" of $M_2(\Bbb R)$. So, $T_A = T_S \circ T_D \circ T_{S^{-1}}$ is a diagonalization of our original $T$.

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All your answer, except for the answer to 1, are valid.

For 1, you only proved that 1 is true for one particular case of $A$, but you have to do it for all values of $A$.

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    $\begingroup$ so, how to prove/disprove 1 ? $\endgroup$ Commented Jul 11, 2017 at 9:04
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Use Kronecker product and the vectorization of a matrix, proof will be obvious.

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For first observe that trace =sum of eigenvalues =a+b=2 And det =ab=-3 So b²-2b+3=0 This will give b=3 or -1 and a=-1 or 3 according to b ...hence u are getting two different eigenvalues that's the reason it is diagonalizable.

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  • $\begingroup$ Could you please explain it? $\endgroup$
    – user444042
    Commented Dec 1, 2017 at 18:53

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