# Diagonalizablility of $T$

Let $M_2(\mathbb R)$ denotes the set of $2\times2$ real matrices. Let $A\in M_2(\mathbb R)$ be of trace $2$ and determinant $-3$. Identifying $M_2(\mathbb R)$ with $\mathbb R^4$, consider the linear transformation $T:M_2(\mathbb R) \to M_2(\mathbb R): B \mapsto AB$. Then which of the followings are true:

(1) $T$ is diagonalizable,

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

(3) $T$ is invertible,

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

Here's how I tried it: Since $0$ is not an eigen value of $A$ so $T$ is so option (3) is correct. To show (2) & (4) are incorrect I considered the matrix $$\begin{pmatrix} 3&0\\ 0&-1\end{pmatrix}\quad$$ which satisfies all the conditions of $A$ & noticed that both $T(B)=2B$ & $T(B)=B$ yeild $B=0$.But I'm clueless about the option (1). Please help.

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Any comment regarding the remaining options will also be appreciated. –  Sugata Adhya Dec 6 '12 at 7:58

Suppose your matrix $A$ is $\left(\begin{matrix} e & f\\g & h\end{matrix}\right).$ Writing $B=( a,b,c,d) \in \mathbb{R}^4$ you can see that $T$ is actually the matrix $$\left(\begin{matrix} e & f & 0 & 0\\ g & h & 0 & 0\\0 & 0 & e & f\\ 0 & 0 & g & h\end{matrix}\right)$$. As $\det(T)=\det(A)^2=9$ we see that $T$ is invertible. Now as $A$ is diagonalizable with eigenvalues -1 and 3, so is $T$, as it is a block matrix: If $P,Q$ are invertible matrices such that $PAQ=\left(\begin{matrix} 3 & 0 \\ 0 & -1\end{matrix}\right)$ you can take $P_1=\left(\begin{matrix} P & 0_2\\0_2 & P \end{matrix}\right)$, $Q_1=\left(\begin{matrix} Q & 0_2\\0_2 & Q \end{matrix}\right)$ where $0_2 =\left(\begin{matrix} 0 & 0 \\0 & 0\end{matrix}\right)$. Then $P_1TQ_1=\left(\begin{matrix} 3 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & 3 & 0\\0 & 0 & 0 & -1\end{matrix}\right)$

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First, notice that the characteristic polynomial of $A$ is $p(\lambda) = \lambda^2 - 2\lambda - 3$ which has two distint real roots so $A$ itself is diagonalizable over $\mathbb{R}$. Now, consider the map $T(B) = AB$. If $T$ has an eigenvalue $\lambda \in \mathbb{R}$, then we have $T(B) = AB = \lambda B$ for some nonzero $B \in M_2(\mathbb{R})$, or $(A - \lambda I)B = 0$.

If $\lambda$ is not an eigenvalue of $A$, then $(A - \lambda I)$ is invertible and so this forces $B = 0$. If $\lambda$ is an eigenvalue of $A$, this forces the image of $B$ to be inside the eigenspace of $A$, a one dimensional vector space.

You have the splitting $$\mathrm{Hom}(\mathbb{R}^2, \mathbb{R}^2) = \mathrm{Hom}(\mathbb{R}^2, \ker(A - 3I) \oplus \ker(A + I)) \cong \mathrm{Hom}(\mathbb{R}^2, \ker(A - 3I)) \oplus \mathrm{Hom}(\mathbb{R}^2, \ker(A + I)).$$

For each of the eigenvalues $\lambda = 3, -1$, you can build two linearly independent matrices which map the two dimensional $\mathbb{R}^2$ into the one dimensional $\ker(A - \lambda I)$, and all four will be linearly independent and diagonalize $T$.

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Both approaches are brilliant. –  Sugata Adhya Dec 6 '12 at 12:31