Ytukyg

I am having a difficult time with the following question. Any help will be much appreciated.

Let $A$ be an $n×n$ real matrix such that $A^T = A$. We call such matrices “symmetric.” Prove that the eigenvalues of a real symmetric matrix are real (i.e. if $lambda$ is an eigenvalue of $A$, show that $lambda = overline{lambda}$ )

$Av=lambda v$ combined with $A=A^T$ gives $$langle Av,Avrangle=v^*A^*Av=v^*A^TAv=v^*A^2v=lambda^2||v||^2.$$

Now $lambda^2=frac{langle Av,Avrangle}{||v||^2}geq 0,$ as the quotient of a nonnegative real number by a positive one. So $lambda$ must be real.

Let $Ax=lambda x$ with $xne 0$, with $lambdainmathbb{R}$, then
begin{align}
lambda bar x^T x &= bar x^T(lambda x)\
&=bar x^T A x \
&=(A^T bar{x})^T x \
&=(A bar x)^T x \
&=(bar A bar x)^T x \
&=(barlambdabar x)^T x\
&=bar lambda bar x^T x.\
end{align}
Because $xne 0$, then $bar{x}^T xne 0$ and $lambda=bar lambda$.

Hint:
Prove that $$x^ast A x=langle x , A xrangle = langle Ax, xrangle = x^ast A^ast x $$
Where $A^ast=overline{A}^T$

If $lambda$ is any eigenvalue of a Hermitian (in particular real symmetric) matrix $A$, then for some non-zero vector $x$,
$$Ax = lambda x implies x^*Ax = lambda x^*x implies lambda = dfrac{x^*Ax}{x^*x}.$$

Now $$lambda^* = dfrac{x^* A^* x}{x^*x} = dfrac{x^*Ax}{x^*x} = lambda.$$
Therefore, $lambda$ is real.

Note: $(AB)^* = B^*A^*$.

Hint: for every $ntimes n$ matrix $M$
$$
langle Mv , wrangle
~=~
langle v , M^H wrangle
$$
where $M^H$ is the conjugate transpose of $M$ and $langle,cdot,,,cdot,rangle$ is the complex inner product (i.e. $langle v,wrangle=v^Hw$).

1. Think about how the eigenvalues of $M^H$ and those of $M$ are related


2. Let $v$ be a $lambda$-eigenvector of $M$ and try what happens choosing $v=w$ in the above equation


3. Now, if $A$ is real valued and symmetric then $A^H=A$; try again the second point with $M=A$...