characteristic polynomial and eigenvalues of $T(A)={ A }^{ t }$

Question

Let $V=M_2(\mathbb R)$ and $T(A)={ A }^{ t }$.

I was asked to find the characteristic polynomial of $T$ and it's eigenvalues, and finally to say if $T$'s diagonalizable.

Is there a way to solve this without actually finding a matrix representation of $T$?

(since I've tried turning it into a set of linear equations and it gets ugly)

Thanks!

Answer 1

You can solve it all in one go. First notice that $T^2 = id$, therefore the only possibles eigenvalues are $\pm 1$. Now you simply need to solve the equations $A^t = A$ and $A^t = -A$. This gives you as eigenvectors for the eigenvalue $1$:

$$\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$$

And for the eigenvalue $-1$: $$\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$$

These four matrices are linearly independent, $M_2(\mathbb R)$ has dimension 4 so they span the whole space. $M_2(\mathbb R)$ has a basis of eigenvectors for $T$, so $T$ is diagonalisable, and its characteristic polynomial is $(X-1)^3(X+1)$. This approach has a straightforward generalization for $M_n(\mathbb R)$ for any $n$.

Answer 2

Since $T^2=\mathrm{Id_{M_2(\mathbb R)}}$ then the polynomial $x^2-1$ annihilates $T$ and since $T\neq\pm \mathrm{Id_{M_2(\mathbb R)}}$ then $x^2-1$ is the minimal polynomial and its roots are simple then $T$ is diagonalizable (we can see that $T$ is diagonalizable since $T$ is a symmetry), moreover by the decomposition $$M_2(\mathbb R)=\mathcal{S}_2(\mathbb R)\oplus \mathcal{A}_2(\mathbb R)$$ we have $1$ is an eigenvalue of $T$ with multiplicity $\dim \mathcal{S}_2(\mathbb R)=\frac{2(2+1)}{2}=3$ and $-1$ is an eigenvalue of $T$ with multiplicity $\dim \mathcal{A}_2(\mathbb R)=\frac{2(2-1)}{2}=1$ so the characteristic polynomial of $T$ is $$\chi(T)=(x-1)^3(x+1)$$

Answer 3

$A^T=\lambda A$ implies that $A=(A^T)^T=(\lambda A)^T=\lambda A^T=\lambda^2A$. So, finding the eigenvalues should be easy. Once the eigenvalues are determined, solve $A^T=\lambda A$ directly to find the eigen "vectors". In this exercise, the eigenvectors should span the whole matrix space and hence the dimension of the eigenspace for each distinct $\lambda$ will tell you the algebraic multiplicity of $\lambda$ in the characteristic polynomial. Finally, $A$ is diagonalisable if and only if its minimal polynomial consists of distinct linear factors.

Answer 4

First think what could be the eigenvalues:

$A^t=\lambda A \implies A=\lambda^2A \implies \lambda=\pm1$

So characteristic polynomial of $T$ will be $P_T(x)=(x-1)^a(x+1)^b$

Any matrix $A$ can be written as sum of a symmetric and a skew symmetric matrix.

You can deduce $a=3$ and $b=1$, and diagonalizability of $T$ follows!