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.

$ T(A) $ can be represented as (double indexes are summed):
$$
[T(A)]_{ij} = T_{ijab} A_{ab} = (\delta_{ai}\delta_{jb}+\delta_{ij}\delta_{2a}\delta_{2b}) A_{ab}
=A_{ij} +A_{22} \delta_{ij}
$$
To find the eigenvalue $\lambda$ you have to assume that there is an eigenvector  $E$:
$$
[T(E)]_{ij} = \lambda   E_{ij} \, ,
$$
which implies (see first equation)
$$
E_{ij} +E_{22} \delta_{ij} = \lambda   E_{ij}
$$
The simplest way to solve this is to have $E_{22}=0$: in this case $E_{ij} +E_{22} \delta_{ij} =E_{ij} = \lambda   E_{ij}$, namely $\lambda =1$.
There is also another eigenvector: if $E_{ij}=0$ for all $(i,j)\neq (2,2)$. In fact, if $i=2,j=2$,
$$
E_{22} +E_{22} \delta_{22} = 2 E_{22}=\lambda   E_{22}
$$
that gives you $\lambda = 2$.
