Show that if $λ$ is an eigenvalue of $M$ then so is $−λ$ I have some trouble with the final steps of a linear algebra proof about eigenvalues.
This is the question as given in the problem:

Let $A$, $B$ and $C$ be $n \times n$ matrices. Suppose that $B$ and $C$ are symmetric. Consider the matrix:
  $$M = \begin{bmatrix}
        A & B \\
        C &-A^T \\
        \end{bmatrix}$$
  Show that if $λ$ is an eigenvalue of $M$ then so is $-λ$.
  Hint: $M$ and $M^T$ have the same eigenvalues

I tried to solve it like this:
$$\det(M-λI) = 0$$
$$\det(M^T -λI) = 0$$
So that means that: $\det(M-λI) = det(M^T-λI)$ 
$$M^T = \begin{bmatrix}
        A^T & C^T \\
        B^T & -A \\
        \end{bmatrix} =\begin{bmatrix}
        A^T & B \\
        C &-A \\
        \end{bmatrix},$$since $B$ and $C$ are symmetric. 
$$\det \begin{bmatrix}
        A-λI & B \\
        C & -A^T-λI \\
        \end{bmatrix} = \det\begin{bmatrix}
        A^T-λI & B \\
        C &-A-λI \\
        \end{bmatrix}$$
$$\det(-AA^T-AλI+λIA^T+λ^2I-BC)=\det(-A^TA-A^TλI+λIA+λ^2I-BC)$$
This can be rewritten to:
$$\det(-AA^T-AλI+λIA^T+λ^2I-BC)=det(-A^TA+AλI-λIA^T+λ²I-BC)$$
This is the point where I am stuck, i see that if you fill in $λ$ on the left side and $-λ$ on the right side, they are the same except for the $-AA^T$ and $-A^TA$. I know that $-AA^T$ and $-A^TA$ are symmetric so they have the same determinant, but does that help in this proof?. 
Is there a way to prove that they are the same? And if not, is there an easier way to do the proof?
 A: Let $(x,y)^T$ be the eigenvector of $M$ which corresponds to the eigenvalue $\lambda$. This means that
$$
\begin{pmatrix}A&B\\C&-A^T\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\lambda x\\\lambda y\end{pmatrix}
$$
or
\begin{align}
Ax+By&=\lambda x,\\
Cx-A^Ty&=\lambda y.
\end{align}
Consider the following multiplication:
$$
M^T\begin{pmatrix}-y\\x\end{pmatrix}=\begin{pmatrix}A^T&C\\B&-A\end{pmatrix}\begin{pmatrix}-y\\x\end{pmatrix}=\begin{pmatrix}-A^Ty+Cx\\-By-Ax\end{pmatrix}=\begin{pmatrix}-\lambda(-y)\\-\lambda x\end{pmatrix}
$$
(the last equation follows from the previous system). Hence, the matrix $M^T$ have eigenvalue $-\lambda$. And since the eigenvalues for $M$ and $M^T$ are the same, the matrix $M$ have eigenvalue $-\lambda$ as well.
A: Alternatively, let $P(x):=\det(M-x\,\text{I})$.  Suppose that $A$, $B$, and $C$ are $n$-by-$n$ matrices.  Then, via multiple column and row swappings, we have
\begin{align}
P(-x)&=\det(M+x\,\text{I})=\det\begin{bmatrix}A+x\,I&B\\C&-A^\top+x\,I\end{bmatrix}\\
&=(-1)^n \,\det\begin{bmatrix}B&A+x\,I\\-A^\top+x\,I&C\end{bmatrix}
\\
&=(-1)^n\,(-1)^n\,\det\begin{bmatrix}-A^\top+x\,I&C\\B&A+x\,I\end{bmatrix}\,.\end{align}
Now, changing the signs of the first $n$ column then the signs of the bottom $n$ rows of $\begin{bmatrix}-A^\top+x\,I&C\\B&A+x\,I\end{bmatrix}$ yields
\begin{align}
P(-x)
&=(-1)^{2n}\,(-1)^n\,\det\begin{bmatrix}A^\top-x\,I&C\\-B&A+x\,I\end{bmatrix}
\\
&=(-1)^{3n}\,(-1)^n\,\det\begin{bmatrix}A^\top-x\,I&C\\B&-A-x\,I\end{bmatrix}
\\
&=(-1)^{4n}\,\det\big(M^\top-x\,I\big)=\det\left((M-x\,I)^\top\right)
\\
&=\det(M-x\,I)=P(+x)\,.
\end{align}
Hence, either the characteristic of the field is $2$ or $P(x)$ is a polynomial in $x$ with only terms of even degrees.  In both cases, $P(\lambda)=0$ iff $P(-\lambda)=0$.  
In fact, in the latter case, it follows that the eigenvalues of $M$ are $\pm\lambda_1,\pm\lambda_2,\ldots,\pm\lambda_n$ for some $\lambda_i$'s in the algebraic closure of the field.  Sergei Golovan's proof shows that the geometric multiplicity of $\lambda$ equals that of $-\lambda$.  My proof shows that they also have the same algebraic multiplicity.
