Eigenvalues and Eigenvectors

Question

For each linear operator $T$ on $V$, find the eigenvalues of $T$ and an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix. Where $V = M_{2\times 2}(\mathbb R)$ and $T \left(\begin{matrix}a & b \\c & d \\\end{matrix}\right) = \left(\begin{matrix}d & b\\ c & a\\ \end{matrix}\right)$ . I just get two eigenvalues both equal to one, the book says there are four eigenvalues, $1$ with multiplicity $3$ and $-1$ with multiplicity $1$. I cannot for the life of me figure out how they got it, please be as detailed as possible.

Answer 1

We first write down the obvious basis for $V$ $$ e_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad e_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad e_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \quad e_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, $$ and then the matrix of $T$ with respect to this basis $$ T = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} $$ by setting $(a,b,c,d)$ equal to $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, and $(0,0,0,1)$ in the given equation for $T$.

Doing cofactor expansion, we see that $\det(T) = -1$. Since the determinant is the product of the eigenvalues, we must have at least one negative eigenvalue! (This is what @Cameron Williams was referring to.)

Since the middle two columns of the matrix of $T$ agree with the identity matrix, we see that $$ \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad\mbox{and}\qquad \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}. $$

If we try multiplying $T$ by $\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T$ and $\begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}^T$, the result is the other vector; by adding, we get $$ \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}. $$ Knowing that we have a negative eigenvalue to find, the next reasonable suspect works: $$ \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix} = - \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ gives us our last eigenvalue of $-1$.

The ordered basis of $V$ we want is then $(e_2, e_3, e_1+e_4, e_1-e_4)$ (or any permutation of it).