What are the eigenvalues of $T(A) = A+ (A)_{2,2}I$, where $T\colon M_2(\mathbb{C})\to M_2(\mathbb{C})$ is a linear operator let $T$ be linear transformation $T:M_2(\mathbb{C})\to M_2(\mathbb{C})$.
$$T(A) = A+ (A)_{2,2}I$$
I need to find its eigenvalues, does it not depend on the value of $(A)_{2,2}$?
I dont understand why $T(A) = 2A$ is the answer (according to the answers the eigenvalue is $2$).
thanks for help.

One approach here is to select a basis, find the $4 \times 4$ matrix of the linear transformation relative to this basis, then find the eigenvalues and eigenvectors of this matrix in the usual way. In this case, however, I prefer to work directly from the definition of eigenvalues and eigenvectors.
Note that $\lambda$ is an eigenvalue of the transformation $T$ if there is a non-zero matrix $A$ for which $T(A) = \lambda A$.  So, we are looking for values of $\lambda$ and $A$ for which the equation
$$
T(A) = \lambda A \implies A + A_{22}I = \lambda A \implies A_{22} I = (\lambda - 1)A.
$$
First, note that if $\lambda = 1$, then any $A$ for which $A_{22} = 0$ is a solution to this equation. Equivalently, we can say that the matrices
$$
\pmatrix{1&0\\0&0}, \quad \pmatrix{0&1\\0&0}, \quad \pmatrix{0&0\\1&0}
$$
form a basis for the eigenspace of $A$ corresponding to $\lambda = 1$.
If $\lambda \neq 1$, then we can rewrite this equation in the form
$$
A = \frac{A_{22}}{\lambda - 1} I.
$$
We can see that two things need to be true in order for this equation to hold. First, $A$ has to be a multiple of the identity matrix, which means that $A = A_{22}I$. With that, this equation becomes
$$
[A_{22}I] = \frac{A_{22}}{(1 - \lambda)}I.
$$
If $A = A_{22}I$ is non-zero, then this can only happen when $1 - \lambda = 1 \implies \lambda = 2$. All together, we see that $T$ has an eigenvalue of $\lambda = 2$, and every associated eigenvalue is a multiple of the identity.
So, the eigenvalues are $1$ (with multiplicity $3$) and $2$ (with multiplicity $1$).
