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Can we say that Eigenvalues of symmetric orthogonal matrix must be $+1$ and $-1$?

Since eigenvalues of symmetric matrices are real and eigenvalues of orthogonal matrix have unit modulus. Combining both result eigenvalues of symmetric orthogonal matrices must be $+1$ and $-1$.

Please clarify whether I am correct? Is there any other approach to solve this problem?

Thanks

Rodrigo de Azevedo
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mathscrazy
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3 Answers3

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Yes, you're right. Also note that if $A^\top A=I$ and $A=A^\top$, then $A^2=I$, and now it's immediate that $\pm 1$ are the only possible eigenvalues. (Indeed, applying the spectral theorem, you can now conclude that any such $A$ can only be an orthogonal reflection across some subspace.)

Ted Shifrin
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Suppose $A$ being symmetric and orthogonal, then we have $A = A^T$ and $A^T A = I$.

Let $\lambda$ be an eigenvalue of $A$. Then we can derive

\begin{align} Ax &= \lambda x \\ A^T A x &= A^T \lambda x \\ x &= A \lambda x \\ \frac{1}{\lambda} x &= Ax = \lambda x\\ \frac{1}{\lambda} &= \lambda \end{align}

So $\lambda$ has to be $\pm 1$.

zyxue
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A symmetric orthogonal matrix is involutory.

Involutory matrices have eigenvalues $\pm 1$ as proved here: Proof that an involutory matrix has eigenvalues 1,-1 and Proving an invertible matrix which is its own inverse has determinant $1$ or $-1$

BCLC
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