# Prove that the eigenvalues of a real symmetric matrix are real.

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

• A real $n\times n$ matrix only can have real eigenvalues (every complex zero of the characteristic is no eigenvalue of the real matrix) – Dominic Michaelis Apr 7 '13 at 19:12
• @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 $C$, $C^* = {\bar {C^T}}$. – Stefan Smith Apr 7 '13 at 19:13
• @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 – Stefan Smith Apr 7 '13 at 19:45
• @StefanSmith if you consider it as a real matrix, it doesn't have any eigenvalues. – Dominic Michaelis Apr 7 '13 at 19:49
• 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. – Julien Apr 7 '13 at 19:59

$$Av=\lambda v$$ 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.

• A quotient of non-negative real numbers? But the argument is very neat. – Chris Godsil Apr 7 '13 at 19:08
• 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 – Dominic Michaelis Apr 7 '13 at 19:19
• 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. – Lepidopterist Apr 7 '13 at 19:23
• 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. – Julien Apr 7 '13 at 19:44
• 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. – Julien Apr 7 '13 at 19:51

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 $$\lambda=\bar \lambda$$.

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

• 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. – Julien Apr 7 '13 at 19:56
• 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 – Dominic Michaelis Apr 7 '13 at 19:59
• 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. – Julien Apr 7 '13 at 20:01

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

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.

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 $M=A$...