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.
 A: It cannot depend on $(A)_{2,2}$, or rather, it does not make sense to say that since $A$ is a variable.

To solve it, let us assume that $M = \begin{pmatrix}a & b\\c & d\end{pmatrix} \in M_2(\Bbb C)$ is an eigenvector (and thus, $M \neq O$) corresponding to eigenvalue $\lambda \in \Bbb C$.
Thus, we have
$$T(M) = \lambda M$$
or
$$M + dI = \lambda M$$
or
$$\begin{pmatrix}a + d & b\\c & 2d\end{pmatrix} = \begin{pmatrix}\lambda a & \lambda b\\\lambda c & \lambda d\end{pmatrix}.$$
Note that the above gives us $4$ equations. These suggest the following cases:


*

*$\lambda = 1$
This forces that $d = 0.$ (Since $\lambda d = 2d.$)
Moreover, if $d = 0$, then we have no further restrictions on $a, b, c$. (Check that the other three equations are always true.)
Thus, this gives us three linearly independent eigenvectors (matrices):
$$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},\;\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix},\;\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}.$$

*$\lambda = 2$
This forces $b = c = 0$.
We also have $a + d = \lambda a = 2a$ and thus, $a = d$.
This gives us the following eigenvector:
$$\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}.$$

*$\lambda \neq 1$ and $\lambda \neq 2$
We don't even have to solve this since we already have four linearly independent eigenvectors and thus, can't possibly get another.
However, even if you do try to solve it, you'll quickly see that this condition will force $a = b = c = d = 0$.

To summarise: You have the following eigenvalues: $2$ and $1$.
You even have the precise eigenspaces.
A: 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 $1$).
A: $ 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 $\lambda = 2$.
A: If $\lambda$ is an eigenvalue of $T$ with eigenvector $A\in M_2(\mathbb{C})$, then we have $T(A)=\lambda A$, hence we get the equations $A_{11}+A_{22}=\lambda A_{11}$,  $2A_{22}=\lambda A_{22}$, $A_{12}=\lambda A_{12}$ and $A_{21}=\lambda A_{21}$. The only solutions to this system are $\lambda=1$ (with $A_{22}=0$) or $\lambda=2$ (with $A_{11}=A_{22}$ and $A_{12}=A_{21}=0$).
A: This linear transformation has eigenvalue $2$ with eigen-element $I$. Notice that
$$
T(I) = I+1I = 2I.
$$
It also has an eigenvalue of $1$ with a 3-dimensional eigenspace spanned by $ \begin{pmatrix}1 & 0 \\0 &0\end{pmatrix}$, $\begin{pmatrix}0 & 1 \\0 &0\end{pmatrix}$, and $ \begin{pmatrix}0 & 0 \\1 &0\end{pmatrix}$, since $T$ acts as the identity map on matrices $A$ with $A_{22}=0$.
