2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Any comment regarding the remaining options will also be appreciated. $\endgroup$ – Sugata Adhya Dec 6 '12 at 7:58
2
$\begingroup$

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 $P_1TQ_1=\left(\begin{matrix} 3 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & 3 & 0\\0 & 0 & 0 & -1\end{matrix}\right)$

$\endgroup$
2
$\begingroup$

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 $T$.

$\endgroup$
  • $\begingroup$ Both approaches are brilliant. $\endgroup$ – Sugata Adhya Dec 6 '12 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.