# Eigenvalue of a symmetric and antisymmetric matrices

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.

http://www.doc.ic.ac.uk/~ae/papers/lecture05.pdf

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.

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