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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?

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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.

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  • $\begingroup$ Nice! I wonder if it's possible to modify your solution for the case when $B$ and $C$ sizes differ (the original problem has the same sizes, but the statement stays if they are different). $\endgroup$ Jan 22 '17 at 17:23
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    $\begingroup$ This way $A$ and $A^T$ are no longer square matrices, so you'd have to make the block expression for the characteristic polynomial more complicated. $\endgroup$ Jan 22 '17 at 17:32
  • $\begingroup$ I checked that the claim does not hold for $A=\begin{bmatrix}1\\2\end{bmatrix}$, $B=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, and $C=1$. So, we need that $A$, $B$, and $C$ be square of the same size. (I forgot that $A-x\,I$ does not make sense when $A$ is non-square.) $\endgroup$ Jan 22 '17 at 17:45
  • $\begingroup$ You're right. My solution works only for the square $A$ too. Sorry for the false alarm. $\endgroup$ Jan 22 '17 at 18:04
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    $\begingroup$ Well, $A$ and $B$ have $n$ columns, and therefore, $n$ pairs of columns have to be swapped. Your attempt is invalid. The determinant of a block matrix $\begin{bmatrix}A&B\\C&D\end{bmatrix}$ is usually not equal to $\det(AD-BC)$. $\endgroup$ Jan 22 '17 at 18:59
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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.

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  • $\begingroup$ Right. I've edited the answer. $\endgroup$ Jan 22 '17 at 16:48
  • $\begingroup$ Thank you for the answer. There is only one thing that I don't understand about your proof: How do you know that you have to pick the eigenvector $(-y,x)^T$ to multiply with $M^T$ so that you get the nice equations that you can substitute. Is there a rule for eigenvectors of a tansposed matrix that you applied? $\endgroup$
    – Zephron
    Jan 22 '17 at 18:43
  • $\begingroup$ I don't know now such a rule. I've just tried a few block vectors starting from $(x,y)^T$ and modifying till something came up. $\endgroup$ Jan 22 '17 at 18:47

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