Determining the eigenvalues for the linear operator $T$ on $V$ $V=M_{2x2}$ 
$T\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}c&d\\a&b\end{bmatrix}; \tag 1$
Attempt: 
I used the standard basis to find
$A = \begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}; \tag 2$ 
then 
$\det(A-\lambda I)=\begin{bmatrix}1-λ&0&0&0\\0&1-λ&0&0\\0&0&1-λ&0\\0&1&0&-λ\end{bmatrix}, \tag 3$
and
$4(1-λ)(1-λ)\begin{bmatrix}1-λ&0\\0&-λ\end{bmatrix}; \tag 4$
so
$(1-λ)(1-λ)(1-λ)(-λ); \tag 5$
In my solutions manual, it says the values of $\lambda$ are $-1,-1,1$ and $1$. I dont understand what Im doing wrong
 A: Note that
$$\det(A-\lambda I)=\begin{vmatrix}-λ&0&1&0\\0&-λ&0&1\\1&0&-λ&0\\0&1&0&-λ\end{vmatrix}=-\lambda(-\lambda^3+\lambda)+(1-\lambda^2)=\lambda^4-2\lambda^2+1$$
A: First, a few remarks on our OP Essie Stern's work; then I'll provide my own take on this one.
Apparently we are using the mapping
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{pmatrix} a \\b \\ c \\ d \end{pmatrix} \tag 1$
to express members of $M_{2 \times 2}$ as column vectors; the map is clearly a linear isomorphism between vector spaces.  It is easy to see that in this representation, the map
$T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ a & b \end{bmatrix} \tag 2$
corresponds to the matrix
$A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}; \tag 3$
the eigenvalues of $A$ then satisfy
$0 = \det(A - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 0 & 1 & 0 \\ 0 & -\lambda & 0 & 1 \\ 1 & 0 & -\lambda & 0 \\ 0 & 1 & 0 & -\lambda \end{bmatrix} \right ); \tag 4$
comparing this to equation (3) in the body of the question indicates that our OP Essie Stein erred in constructing the matrix $A - \lambda I$, apparently replacing some of the $1$s in off-diagonal positions with $1$s on the diagonal, which then give rise to the $1 - \lambda$ entries occurring in (3) of the question.  
To continue, we may expand (4) in minors along the first column, yielding
$\det(A - \lambda I) = -\lambda \det \left ( \begin{bmatrix} -\lambda & 0 & 1 \\ 0 & -\lambda & 0 \\ 1 & 0 & - \lambda \end{bmatrix}  \right ) + 1 \cdot \det \left (\begin{bmatrix} 0 & 1 & 0 \\ -\lambda & 0 & 1 \\ 1 & 0 & -\lambda \end{bmatrix} \right )$
$= -\lambda(-\lambda^3 + \lambda) + 1 \cdot (1 - \lambda^2) = \lambda^4 - 2 \lambda^2 + 1 = (\lambda^2 - 1)^2; \tag 5$
the roots of
$(\lambda^2 - 1)^2 = 0 \tag 6$
are easily seen to be $-1, -1, 1, 1$; these then are the eigenvalues of the $4 \times 4$ matrix $A$; it is from here but a short step to find eigvectors corresponding to these values by solving the equation
$\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}\begin{pmatrix} a \\b \\ c \\ d \end{pmatrix} = \begin{pmatrix} c \\ d \\ a \\ b \end{pmatrix} = \pm 1 \begin{pmatrix} a \\b \\ c \\ d \end{pmatrix}; \tag 7$
we in fact must have
$(a, b) = \pm (c, d); \tag 8$
it is from here but a short step to see that the $+1$ and $-1$ eigenspaces of $A$, and hence $T$, are each two-dimensional, and to work out which $2 \times 2$ matrices lie in each eigenspace; the details are straighforward and left to the reader.
Having shown how the approach suggested by our OP Ellie Stein is supposed to work, I will now provide the promised simpler solution:
We observe from (2) that
$T^2\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}; \tag 9$
thus
$T^2 = I \tag{10}$
on $M_{2 \times 2}$; it follows then that the eigenvalues of $T$ must satisfy
$\lambda^2 - 1 = 0, \tag{11}$
that is,
$\lambda \in \{-1, 1 \}; \tag{12}$
the eigenvector matrices of $T$ must then obey
$T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ a & b \end{bmatrix} = \lambda \begin{bmatrix} a & b \\ c & d \end{bmatrix}; \tag{13}$
it is then easy to see by inspecting (13) that we must have
$\lambda = 1 \Longleftrightarrow (c, d) = (a, b), \tag{14}$
$\lambda = -1 \Longleftrightarrow (c, d) = -(a, b), \tag{15}$
which together show that both $-1$ and $1$ are in fact eigenvalues of $T$; the eigenspaces of each are in fact two-dimainsional, being in fact matrices of the form 
$\begin{bmatrix} a & b \\ a & b \end{bmatrix} \; \text{when} \; \lambda = 1, \tag{16}$
and
$\begin{bmatrix} a & b \\ -a & -b \end{bmatrix} \; \text{when} \; \lambda = -1. \tag{17}$
