10
$\begingroup$

I know that if $A=A^T$ is real, then the eigenvalues are real, and that eigenvectors corresponding to different eigenvalues are orthogonal. Is there an easy way to see what happens in the case of a repeated eigenvalue, or does this require some sophisticated arguments (that I guess will come later in my book)?

$\endgroup$
6
  • 4
    $\begingroup$ The relevant result is called the spectral theorem for symmetric matrices and is not entirely straightforward. It states that symmetric matrices are diagonalizable and as a corollary, you see that if some eigenvalue has algebraic multiplicity $k$ then it must come with $k$ linearly independent eigenvectors. $\endgroup$
    – levap
    Commented Nov 7, 2015 at 16:36
  • $\begingroup$ start comparing two diagonals $\left(\begin{array}{cc}a&0\\0&a\end{array}\right)$ versus $\left(\begin{array}{cc}b&0\\0&c\end{array}\right)$ with $b\neq c$ $\endgroup$
    – janmarqz
    Commented Nov 7, 2015 at 16:41
  • 1
    $\begingroup$ @levap thanks for your pointer to the theorem -- it helps a lot! So it's right to say for each eigenvalue of symmetric matrices, the algebraic multiplicity and geometric multiplicity are always equal. Am I right? $\endgroup$ Commented Jul 25, 2016 at 3:47
  • 1
    $\begingroup$ @SibbsGambling: Yep, that is correct. $\endgroup$
    – levap
    Commented Jul 25, 2016 at 3:49
  • $\begingroup$ Consider the case $A = 1$ for comparison. $\endgroup$
    – anomaly
    Commented Aug 23, 2018 at 19:47

1 Answer 1

5
$\begingroup$

The best argument I know of is actually completely agnostic about whether the eigenvalues are repeated. It depends on the following simple calculation:

Suppose $x$ is an eigenvector of $A$, $A$ is symmetric, and $y$ is orthogonal to $x$. Then

$$(x,Ay)=(Ax,y)=(\lambda x,y)=\lambda (x,y)=0.$$

Thus $Ay$ is also orthogonal to $x$.

Now we know that if $x$ is an eigenvector of $A$, then the orthogonal complement of the span of $x$ is an invariant subspace under $A$. In other words, the restriction of $A$ to the orthogonal complement of the span of $x$ is an endomorphism (a linear map from a vector space to itself). Since everything is finite dimensional, this map must have an eigenvalue. You can also show that this restriction is symmetric again. But now we can do the procedure again, reducing the dimension by $1$ at each step, until we get to dimension $1$ where the result is trivial.

An interesting feature of this argument is that we do not extract the entire eigenspace of the eigenvalue $\lambda$ at once. This means we do not have to directly prove that the algebraic multiplicity of $\lambda$ and the geometric multiplicity of $\lambda$ coincide. Instead, maybe we get that eigenvalue again during the construction, maybe we don't. The procedure doesn't care either way.

Incidentally, in the case of a repeated eigenvalue, we can still choose an orthogonal eigenbasis: to do that, for each eigenvalue, choose an orthogonal basis for the corresponding eigenspace. (This procedure does that automatically.) The only funny thing about this case is that there exist non-orthogonal eigenbases.

$\endgroup$
7
  • $\begingroup$ Good answer, but also found this answer to be quite lucid: quora.com/… $\endgroup$ Commented Mar 16, 2020 at 23:27
  • $\begingroup$ @Ian: I don't understand how you can guarantee there must exist an eigenvalue in the orthogonal complement of the span of $x$. Since we only know there exist at least one eigenvector in $R^n$. $\endgroup$
    – Andrew
    Commented Jun 7, 2022 at 16:09
  • $\begingroup$ @Andrew It is specifically because that orthogonal complement is invariant under $A$, which is what the previous calculation showed (using the fact that $A$ is symmetric). This lets you view $A$ as restricted to the orthogonal complement of the span of $x$ as an endomorphism of complex vector spaces in its own right, and every endomorphism of finite dimensional complex vector spaces has an eigenvalue. (You need to clean this up a little bit to get the $\mathbb{R}$ analogue.) $\endgroup$
    – Ian
    Commented Jun 7, 2022 at 16:15
  • $\begingroup$ @Ian Sorry but I only know the result which I mentioned in my comment. So maybe what you said is a generalization of that result? $\endgroup$
    – Andrew
    Commented Jun 7, 2022 at 16:37
  • $\begingroup$ @Andrew The proof is slightly more complicated but the intuition is the same: ultimately you just identify the orthogonal complement of the span of $x$ with $\mathbb{R}^{n-1}$. But the key thing is that this space is invariant, and it may not be completely obvious why that's important (why can't you just change basis in the domain and change basis in the codomain and have that just work?) $\endgroup$
    – Ian
    Commented Jun 7, 2022 at 16:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .