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What condition, if any, can be said about a real symmetric matrix to have all its eigen values appear in pairs with opposite sign but same magnitude, i.e. if $\lambda$ is an eigen value of A then $-\lambda$ should also be an eigen value of A?

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    $\begingroup$ An obvious necessary condition is a vanishing trace. The characteristic polynomial is of the form $(x^2-a^2)(x^2-b^2)...$ being another one. $\endgroup$ – Tapu Mar 23 '13 at 6:13
  • $\begingroup$ Somebody down voted the question. May I know the reason for that. $\endgroup$ – Anurag A Jan 8 '16 at 23:24
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Here are some necessary and sufficient conditions:

  • $A$ is $n\times n$ for some even $n$ and it has the Jordan form $\lambda_1D\oplus\lambda_2D\oplus\cdots\lambda_{n/2}D$, where $\lambda_1,\lambda_2,\ldots,\lambda_{n/2}$ are nonzero real numbers and $D=\operatorname{diag}(1,-1)$. Hence its characteristic polynomial is $\prod_{k=1}^{n/2}(x^2-\lambda_k^2)$.
  • $A$ is nonsingular and similar to $-A$.
  • $A$ is similar to a block matrix of the form $\begin{pmatrix}0&B\\ B&0\end{pmatrix}$ where $B$ is nonsingular and real symmetric.
  • $A$ is similar to a block matrix of the form $\begin{pmatrix}B&0\\ 0&-B\end{pmatrix}$ where $B$ is nonsingular and real symmetric.

For necessary conditions, three obvious ones are $A$ is traceless, $A$ is nonsingular and $A$ has an even order.

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Lets see if we can develop the very nice comment by Tapu.

If we look at the 2x2 case for real symmetric matrices, we have:

$$\begin{bmatrix}a & b \\ b & d\end{bmatrix}$$

Writing out and solving the characteristic polynomial of $|A -\lambda I|=0$, and solving for the the eigenvalues, we get:

$$\displaystyle \lambda_1 = \frac{1}{2} (-\sqrt{a^2-2 a d+4 b^2+d^2}+a+d)$$

$$\displaystyle \lambda_2 = \frac{1}{2} (+\sqrt{a^2-2 a d+4 b^2+d^2}+a+d)$$

We want the condition $\lambda_1 = \lambda_2$.

For this, we get $b = 0 ~\text{and}~ a = d$. However, we have the added condition that we want the eigenvalues to be opposite sign, so we will setup the real symmetric matrix as:

$$\begin{bmatrix}d & 0 \\ 0 & -d\end{bmatrix}$$

What do you notice about the lower and upper triangular? What do you notice about the trace? What do you notice about the characteristic polynomial?

Using those observations, lets see if we can write out the $4x4$ case and if this generalizes.

$$\begin{bmatrix}a & 0 & 0 & 0 \\ 0 & -a & 0 & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & -b\end{bmatrix}$$

It is real symmetric. The upper and lower triangular is zero. The characteristic polynomial is $(a-\lambda) (a+\lambda) (b-\lambda) (b+\lambda)$, that is $(\lambda^2 -a^2)(\lambda^2-b^2)$, so we get eigenvalues as $\pm a, \pm b$. Look at the trace of our matrix again.

Now, can you use these observations and make an argument for the general case?

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  • $\begingroup$ Nice observation Amzoti $\endgroup$ – mrs Mar 24 '13 at 10:26
  • $\begingroup$ Really nice work! $\endgroup$ – Namaste Apr 16 '13 at 1:02
  • $\begingroup$ @amWhy: Thanks - I enjoyed this question. $\endgroup$ – Amzoti Apr 16 '13 at 1:04

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