Question on finding eigenvalues of another linear transformation

I have a quick question on finding eigenvectors for a linear transformation. I'm given:

$T(A) = A^t$ where $A = M_2$ i.e. a $2 \times 2$ matrix consisting of real numbers.

So the general approach is to solve $T(v) = \lambda v$, and from there I set up:

\left\{\begin{align*} a &= \lambda a \\ d &= \lambda d \\ b &= \lambda c \\ c &= \lambda b \end{align*}\right.

which suggests that that $\lambda = 1$ is the only eigenvalue. Given that, I can verify that $T(v) = \lambda v$ holds when $\lambda = 1$ and $b = c$.

But I'm not sure how to describe the eigenvector. How do I put it into terms? I was thinking that a basis for the solution set would be:

$$a \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\\ 1 & 0 \end{pmatrix}$$

but that's not an eigenvector. How exactly would I describe the eigenvector for $\lambda = 1$?

Edit: So if you have a matrix with two eigenvectors, both of which are 2 x 2 matrices, how would you go about diagonalizing it? i.e. if you have to solve $\Lambda = SAS^{-1}$, how do you determine S?

• Why do you say that the matrix you've written is not an eigenvector? The eigenvectors for the eigenvalue $1$ are the symmetric matrices. – Davide Giraudo Dec 13 '11 at 16:59
• Oh okay, I didn't realize that. So is 1 the only eigenvalue as well, or are there other solutions? Also, let's say if we later wanted to diagonalize A, but had an "eigenelement," how would we derive S? Normally I just put the eigenvectors in, but if the eigenvectors were 2x2 matrices, would S have blocks of eigenvectors? – John Doe Dec 13 '11 at 17:05
• $-1$ is also an eigenvalue, with eigenvector the skew-symmetric matrices. – Davide Giraudo Dec 13 '11 at 17:11
• Ahh, that makes sense. Is there a systematic way to find eigenvalues once you have your equations, or is it mostly check specific values and see what happens? – John Doe Dec 13 '11 at 17:14
• It's worth pointing out that generally, for small dimensional vector spaces like this one, it is more difficult to try to solve the equation $T(\mathbf{v})=\lambda\mathbf{v}$ simultaneously for $\lambda$ and for nonzero $\mathbf{v}$, than to simply go ahead, find a basis for $\mathbf{V}$, determine the matrix of $T$ relative to the basis, and then use the characteristic polynomial of the matrix to find the eigenvalues. – Arturo Magidin Dec 13 '11 at 17:31

It seems that you're confused by the two different uses of the term "vector". In one use, a vector is a column or row of numbers, a matrix one of whose dimensions is $1$. In another use, a vector is any element of any vector space. It is this second sense of the term that's being used when we say that the matrices you describe in your last displayed equation are the eigenvectors corresponding to the eigenvalue $1$ of this linear transformation.

The equation $\Lambda = SAS^{-1}$ doesn't make sense here since you're examining the eigenvalues and eigenvectors of $T$, not of $A$, so you want $\Lambda = STS^{-1}$. You can choose a basis for the vector space of $2\times2$ matrices, for instance the canonical basis, and then express $T$ as a $4\times4$ matrix with respect to this basis. If you also express the eigenvectors you found (three for $1$ and one for $-1$) in this basis, you'll find that they are eigenvectors, in the first sense of the word, of the $4\times4$ matrix representing $T$, and that a $4\times4$ matrix formed out of them diagonalizes the matrix representing $T$.

• I didn't realize that a eigenvector can be bigger than $n x 1$. I did edit my question in the end though, so if you could take a look at that, I'd appreciate it! – John Doe Dec 13 '11 at 17:11

Arturo and joriki gave you good advices. You should find the matrix of your endomorphism $T(A) = A^t$ associated to the standard basis of $M_2(\mathbb{R})$:

$$E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \ , \qquad E_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \ , \qquad E_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \ , \qquad E_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \ .$$

That is,

$$T(E_1) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \ , \qquad T(E_2) = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \ , \qquad T(E_3) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \ , \qquad T(E_4) = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \ .$$

From which you immediatly realise that there is at least one eigenvalue, $1$, with at least a $2$-dimensional subspace of eigenvectors -the span of $E_1$ and $E_4$. To get everything, you just put this data into the associated matrix of $T$:

$$M = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

compute its characteristic polynomial:

$$Q_M(t) = (t-1)^3 (t+1)$$

and find out that there was an eigenvalue you didn't see: $-1$.

If you compute now the respective eigenspaces, you'll find that

$$\ker (M - I ) = \mathrm{span} (E_1, E_2 + E_3, E_4)$$

is the subspace of symmetric matrices and

$$\ker (M + I) = \mathrm{span}(E_2 - E_3)$$

is the subspace of skew-symmetric matrices.