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T is linear transformation working on 2x2 matrices:

T(A) = $\begin{bmatrix}1 & 1\\1 &1\end{bmatrix}$ A

as far as I see only 0 is an eigen value but someone told me 2 is eigen value too and I can't understand why , can anyone help please?

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Because $\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix} = 2 \begin{bmatrix}1\\1\end{bmatrix}$. –  Algebraic Pavel Apr 4 at 16:19

3 Answers 3

$$T\left(\begin{bmatrix}1&1\\1&1\end{bmatrix}\right)=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}=\begin{bmatrix}2&2\\2&2\end{bmatrix}=2\begin{bmatrix}1&1\\1&1\end{bmatrix}$$

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Using the general method to find eigenvalues (and the characteristic polynomial of a matrix, of course):

$$\det(\lambda I-A)=\begin{vmatrix}\lambda-1&-1\\ -1&\lambda-1\end{vmatrix}=\lambda^2-2\lambda=\lambda(\lambda-2)=0\iff \lambda=\begin{cases}0\\2\end{cases}$$

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Since $T$ is a linear map $\mathbb R^4\to\mathbb R^4$ shouldn't we be looking at a polynomial of degree $4$? –  Christoph Apr 4 at 16:41
    
Probably perhaps of the poor formating but I didn't see that: I thought it was a map on two dimensional space. Yes, you're right, yet the basics remain the same. –  DonAntonio Apr 4 at 16:43
    
In any event the problem reduces to the ordinary matrix-vector case $T \vec v = \lambda \vec v$; see my answer. –  Robert Lewis Apr 4 at 20:58

Well, $0$ is an eginvalue since

$T(\begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix}) = 0, \tag{1}$

as is most easily seen. Likewise we have

$T(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}) = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2 \begin{bmatrix} 1 &1 \\ 1 & 1 \end{bmatrix}, \tag{2}$

so $2$ is also an eigenvalue. So far we have proceeded by intuition an (hopefully intelligent) guesswork, but we can sytematically find all the eigenvalues and eigenvectors of $T$ by observing that, if we write $A$ in columnar form, that is,

$A = [\vec a_1 \; \vec a_2 ], \tag{3}$

where $\vec a_1$ and $\vec a_2$ are the columns of $A$, then

$TA = [T\vec a_1 \; T \vec a_2], \tag{4}$

so that $TA = \lambda A$ becomes

$TA = [T\vec a_1 \; T \vec a_2] = \lambda [\vec a_1 \; \vec a_2 ] = [\lambda \vec a_1 \; \lambda \vec a_2]; \tag{5}$

(5) shows that the "eigenmatrices" of the operator $T$ associated with eigenvalue $\lambda$ are precisely those non-zero matrices $A$ whose non-zero columns are ordinary $\lambda$-eigenvectors of $T$. Thus we may find the eigenvalues of $T$ as it operates on $2 \times 2$ matrices by simply finding the ordinary eigevalues of $T$ as it maps vectors to vectors. Thus we construct the characteristic polynomial $p_T(\lambda)$ of $T$:

$p_T(\lambda) = \det (T - \lambda I) = \det(\begin{bmatrix} 1 - \lambda & 1 \\ 1 & 1 - \lambda \end{bmatrix}) = (1 - \lambda)^2 - 1 = \lambda^2 - 2\lambda; \tag{6}$

we see the roots of $p_T(\lambda)$ are $0$ and $2$, as indicated by our intuitive approach. However, (6) shows that $0, 2$ are the only eigevalues of $T$, whether it is applied to vectors or matrices. And the "eigenmatrices" may be constructed from the ordinary eigenvectors of $T$; since the $0$-eigenvector of $T$ is $\vec e_0 = (1, -1)^T$ and the $2$-eigenvector is $\vec e_2 = (1, 1)^T$, we easily see that the matrices

$[\vec e_0 \; 0], [0 \; \vec e_0] \tag{7}$

are a basis for the $2$-dimensional $0$-eigenspace of $T$ as it applies here to $2 \times 2$ matrices; likewise

$[\vec e_2 \; 0], [0 \; \vec e_2] \tag{8}$

form a basis for the $2$-eigenspace; our "guesswork" matrices are simply the sums of the matrices in (7) and (8) respectively; each eigenspace of $T$ as applied to matrices is two-dimensional, in a sense composed of two copies of the corresponding eigenspaces of $T$ as applied to vectors. In this sense the answer given by DonAntonio is correct in spirit, for it acknowledges, though tacitly, that the evaluation of eigenvalues in fact reduces to, and is determined by, the action of $T$ on simple vectors.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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