# Is this a valid proof of eigenvectors/values of nonzeroilty?

Let $A$ be a square real matrix such that it is symmetric, show that if $Ax = \lambda x$ for some nonzero vector in $\mathbb{C}^n$, then in fact, $\operatorname{Re}(x)$ is an eigenvector of $A$ if it is nonzero given that $\lambda$ is real.

I thought that to prove that the real part of $x$ is real, I would do a contrapositive and a contradiction that is, "If $\mathbf{x}$ is non-zero, then \operatorname{Re}(x) is nonzero" will turn to "If $\operatorname{Re}(x)$ is zero, then $x$ is zero".

Proof

Fact: Eigenvectors are by definition not zero vectors

Assume that $\operatorname{Re}(x)$ is zero, then $\mathbf{x}$ is zero.

So we have $\mathbf{x} = \operatorname{Re}(x) + \operatorname{Im}(x)i = 0 + \operatorname{Im}(x) = \operatorname{Im}(x)i$

and we arrive at $\mathbf{x} = \operatorname{Im}(x)i$, however $\mathbf{x} \neq 0$, so $\operatorname{IM}(x) \neq 0$, therefore this is a contradiction and $\operatorname{Re}(x)$ is nonzero.

Q.E.D

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You haven't used the hypothesis that $A$ is symmetric, which makes me wonder why it's there. – Gerry Myerson Jan 29 '12 at 23:11

Since $A$ is real $\overline{Ax}=A\bar x$ and by the properties of inner product $\lambda$ has to be a real number. So $A\bar x=\overline{Ax}=\overline{\lambda x}=\lambda\bar x$ and so $$A(\operatorname{Re}x)=A\left(\frac{x+\bar x}2\right)=\frac 12(Ax+A\bar x)=\frac 12(\lambda x+\lambda \bar x)=\lambda\operatorname{Re} x,$$ hence if $\operatorname{Re}\neq 0$ it's an eigenvector for $\lambda$.
But $\operatorname{Re}x$ can be $0$, for example taking $A=Id$, $\lambda=1$ and $x=i(1,0)$, then $Ax=x$ but of course $\operatorname{Re} x=0$.
Identity matrix (I could take the $0$ matrix but it's a degenerate example). – Davide Giraudo Jan 29 '12 at 20:14
It's $i$ times the first vector of the canonical basis of $\mathbb C^2$. It's the statement $x\neq 0$ implies $\operatorname{Re}x\neq 0$ which is wrong. – Davide Giraudo Jan 29 '12 at 20:23
Sorry i have no idea what a canonical basis is. I proved that $Re(x) \neq 0$ implies $x \neq 0$, not $x \neq 0$ implies $Re(x) \neq 0$ – lem Jan 29 '12 at 20:25