Find the eigenvectors and eigenvalues of a matrix linear transformation

Question

So I've been given this in college :

Find all the eigenvectors and eigenvalues of the linear transformation $\textsf T : \textsf{M}_{2 \times 2}(\mathbb R) \to \textsf{M}_{2 \times 2}(\mathbb R)$ given by

$$\textsf{T}(M) = M + (aM)^t$$

($a$ being a real number).

Once done, find the $a$ values which make $\textsf T$ non-diagonizable.

Thanks and sorry if its hard to understand. I'm new to posting on this site and I'm still working on my english.

Answer 1

Think about a matrix that upon left-multiplying with $M$ would give $M$ and another matrix that also upon left-multiplying with $M$ would give $a$ times the transpose of $M$, $M^t$. The transformation matrix is the sum of these two matrices.

Then for this matrix to be diagonalizable, the multiplicity of each eigenvalue must match the number of corresponding eigenvectors.

Answer 2

if you map the matrix $\begin{pmatrix} x & y \\ z & w \end{pmatrix}$ to the vector $\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}$ then you may represent $T$ by:

$$ \begin{pmatrix} 1 + a & 0 & 0 & 0 \\ 0 & 1 & a & 0 \\ 0 & a & 1 & 0 \\ 0 & 0 & 0 & 1+ a \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} $$

which gives three eigenvalues equal to $1+a$ and one to $1-a$. the, former corresponding to the symmetric 2 x 2 matrices as eigenvectors, and the latter to scalar multiples of $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. as to special eigenvalues you might want to begin by looking at $a \in \{0,\pm 1\}$.