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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 $\lambda = \overline{\lambda}$ )

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  • $\begingroup$ A real $n\times n$ matrix only can have real eigenvalues (every complex zero of the characteristic is no eigenvalue of the real matrix) $\endgroup$ – Dominic Michaelis Apr 7 '13 at 19:12
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    $\begingroup$ @Susan : see Dominic's answer. You will need to use the "complex inner product" $\langle \mathbf{x}, \mathbf{y} \rangle = \sum_{i=1}^n {\bar x_i}y_i$. Also see Lepidopterist's answer, where $C^*$ is the conjugate transpose of $C$, $C^* = {\bar {C^T}}$. $\endgroup$ – Stefan Smith Apr 7 '13 at 19:13
  • $\begingroup$ @DominicMichaelis : do you really mean that (a real square matrix can have only real eigenvalues)? I'm afraid you might confuse Susan. What about $[0, 1;-1, 0]$ with eigenvalues $\pm i$? (Sorry, I don't remember the $\LaTeX$ for writing a matrix $\endgroup$ – Stefan Smith Apr 7 '13 at 19:45
  • $\begingroup$ @StefanSmith if you consider it as a real matrix, it doesn't have any eigenvalues. $\endgroup$ – Dominic Michaelis Apr 7 '13 at 19:49
  • $\begingroup$ Going Dominic's way or Lepidopterist's way, you will easily see that the (a priori complex) eigenvalues must be real. Note that the exact same proofs show that the eigenvalues of a hermitian $A^*=A$ matrix are real in the general complex case. $\endgroup$ – Julien Apr 7 '13 at 19:59
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$Av=\lambda v$ combined with $A=A^T$ gives $\langle Av,Av\rangle=v^*A^*Av=(Av)^*Av=(\lambda v)^*(\lambda v)=\lambda\lambda^*v^*v=\lambda\lambda^*||v||^2=|\lambda|^2||v||^2.$

where $|\lambda|$ is the complex modulus of $\lambda$.

$v^*a^2v$ does NOT equal $\lambda^2||v||$ in the general case. It equals $|\lambda|^2||v||^2$ as it equals $\lambda\lambda^*||v||$ as shown above. Therefore this method does not actually show the result.

Other answers to the question are correct.

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  • $\begingroup$ A quotient of non-negative real numbers? But the argument is very neat. $\endgroup$ – Chris Godsil Apr 7 '13 at 19:08
  • $\begingroup$ only because of $\lambda^2$ is real $\lambda$ is not real in general. and still you need non negative, because $Av=0$ will surely not be positive. and your equation is confusing $\endgroup$ – Dominic Michaelis Apr 7 '13 at 19:19
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    $\begingroup$ If $\lambda^2$ is positive and real, then $\lambda$ is real. What is confusing? And ok, we can assume $\lambda$ is not zero! Zero is real obviously. $\endgroup$ – Lepidopterist Apr 7 '13 at 19:23
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    $\begingroup$ You don't need to bother about $\lambda =0$. All you need is $\|v\|>0$, which is given as you take a (nonzero) eigenvector. So you do find $\lambda^2\geq 0$. Which is equivalent to $\lambda $ being real. $\endgroup$ – Julien Apr 7 '13 at 19:44
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    $\begingroup$ I decided to edit your answer directly to avoid long chat discussions. I hope you don't mind. Just roll back to your version if you disagree and I will leave it alone. $\endgroup$ – Julien Apr 7 '13 at 19:51
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Let $Ax=\lambda x$ with $x\ne 0$, with $\lambda\in\mathbb{R}$, 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 \\ &=(\bar\lambda\bar x)^T x\\ &=\bar \lambda \bar x^T x.\\ \end{align} Because $x\ne 0$, then $\bar{x}^T x\ne 0$ and $\lambda=\bar \lambda$.

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

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    $\begingroup$ As $A^*=A$, you even get $x^*A^*x=x^*Ax$ directly without any intermediate step. ANd that's all you need for nonzero eigenvalues to be real. I have just seen you hint, by the way, +1. $\endgroup$ – Julien Apr 7 '13 at 19:56
  • $\begingroup$ It's homework so I thought i only give a hint. yeah the equation is pretty obvious but it's worth mentioning in homework i guess $\endgroup$ – Dominic Michaelis Apr 7 '13 at 19:59
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    $\begingroup$ I'm afraid you misunderstood my comment. It is great that you only gave a hint. I was just saying that the first two equalities are not necessary. But sure, it does not hurt either to include them. $\endgroup$ – Julien Apr 7 '13 at 20:01
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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: $(AB)^* = B^*A^*$.

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Since $A^* = A$, $(x^*Ax)^* = x^*Ax$. Therefore $x^*Ax$ is a real number for any $x$. If $x$ is an eigenvalue of $A$ with eigenvalue $\lambda$, we have $x^*Ax = x^*(\lambda x) = \lambda x^*x$. Since $x^*Ax$ and $x^*x$ are always real (and $x^*x$ is not zero for an eigenvector $x$), this means $\lambda$ must be real too.

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

  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 $M=A$...
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