Let A be a real n×n matrix;

1) If A is symmetric, show that all eigenvalues of A (in complex numbers) are real.

2) If A is antisymmetric, show that all eigenvalues of A are pure imaginary. (They are like b$i$ where b is a real number and $i^2 = -1$).

I found an article related, but I don't understand it.



1 Answer 1


Question 1) is very classical.

Question 2: Let ($\lambda, V$) be an eigenpair of $A$, i.e., be such that $$AV=\lambda V \ \ \ (1)$$.

As $A^T=-A$, $A^TA=-A^2$. Thus, using (1) twice

$$(A^TA)V=-AAV=-A(\lambda V)=-\lambda AV=-\lambda^2 V$$.

Therefore $-\lambda^2$ is an eigenvalue of $A^TA$.

But, for any $A$, $A^TA$ is a symmetric matrix ($(A^TA)^T=A^TA^{TT}=A^TA$).

that is in fact with positive semi definite I. E. with $ \geq 0$ eigenvalues. Thus, according to question 1), $A^TA$ has real eigenvalues. Thus, $-\lambda^2 \geq 0 \iff \lambda^2 \leq 0$. Therefore $\lambda=ib$ where $b$ is a real number.

Another method of proof relies on quadratic (hermitian) forms : see here.

  • $\begingroup$ Can you please give a hint on question 1? $\endgroup$ Commented May 22, 2016 at 14:43
  • $\begingroup$ Hers is one : (www.quandt.com/papers/basicmatrixtheorems.pdf) $\endgroup$
    – Jean Marie
    Commented May 22, 2016 at 17:25

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