complex and real spectral theorem for matrices

Question

I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $\mathbb{R}$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.

Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $\mathbb{C}$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?

Thanks in advance.

Answer 1

Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $\lambda$. Note that \begin{align} \lambda \overline{v}=\overline{\lambda v} = \overline{Av}= A\overline{v} \end{align} then $\overline{v}$ is also an eigenvector. Hence $v+\overline{v}$ is a real eigenvector. So pick this one.

Answer 2

Because if $\lambda$ is an eigenvalue of $A$ and $\lambda\in\mathbb R$, then $A$ has an eigenvector in $\mathbb{R}^n$ whose corresponding eigenvalue is $\lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $\lambda$ is equivalent to asserting that $\lambda$ is a real root of the characteristic polynomial of $A$.

Answer 3

It's mostly an addition to Jacky Chong's answer. Even though he left a comment about why the real eigenvectors constructed by taking $v + \bar v$ are ortgogonal, I really fail to see how taking the real component helps. I mean it's easy to come up with orthogonal complex vectors such that their real components aren't orthogonal: $(1, i)$ and $(1, -i)$, for example. Maybe I'm just missing his point. So here's my proof. Let $v$ and $w$ be eigenvectors with $\lambda$ and $\mu$ as corresponding eigenvalues (here $\lambda \neq \mu$) Then $$ \lambda \langle v, w \rangle = \langle \lambda v, w \rangle = \langle Av, w\rangle = \langle v, A^T w \rangle = \langle v, Aw \rangle = \langle v, \mu w \rangle = \mu \langle v, w \rangle $$ If $\langle v, w \rangle \neq 0$, then we can divide by it and get $\lambda = \mu$, so it must be zero.

Answer 4

I am afraid that we may not be able to get the diagonalizability of real symmetric matrices (with real matrices) from the complex spectrum theorem. The reason is simply that even if two $\mathbb C^n$ vectors $u,v$ are orthogonal w.r.t the complex inner product, their real parts $Re(u)$ and $Re(v)$ may not be orthogonal w.r.t the real inner product. One example will be the one provided by @Павло Сурженко by letting $u=(1,i)$ and $v=(1,-i).$ Note that in this example, we even have that $Im(u)\not\perp Im(R)$ w.r.t the real inner product.

So basically there is NO way to get the diagonalizability of real symmetric matrices from the complex spectrum theorem if you require the matrices which are used to diagonalize real symmetric matrices to be real.

However, you can prove the diagonalizability of real symmetric matrices straightly and the proof is basically the same with the proof of the diagonalizability of complex Hermitian matrices(which are the complex generalization of real symmetric matrices).

The reason why we can prove the diagonalizability of real symmetric matrices directly is that we can at the very beginning choose real eigenvectors. We can not do the other way around, i.e. first picking up complex eigenvectors and later setting them all real after proving the spectrum theorem.