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If $A$ is symmetric (real) matrix,
CASE 1: $\lambda$ distinct $\rightarrow$ eigenvectors are orthonormal
CASE 2: $\lambda$ not distinct $\rightarrow$ eigenvectors are orthogonal (and then they can be normalized)

I'm focusing on CASE 2.
It is trivial becuase every symmetric matrix can be diagonalized by orthogonal matrix.
But I want to show that mathematically. (eg. inner product, etc...)
Also, in CASE 2, the number of eigenvectors are equal to multiplicity of eigenvalues. Why they can not exceed that? (It's trivial too, but I want to show it precisely!)
We can conclude this by constructing an orthonormal basis of $R^n$ using eigenvectors of $A$.

I think I know principal theorems related to them but have problems with applying them.
Please help me and don't just throw a hint, I need detailed explanation!
Thanks.

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    The key facts here are that the matrix can be diagonalized, and all eigenvalues are real. So, try to prove the results above using a diagonal matrix. Remember, eigenvectors are particularly simple with diagonal matrices. – copper.hat Sep 15 '13 at 17:03
  • @copper.hat Thanks for your comment. But as I mentioned above, I tried to find the answer using every theorem and calculation but failed. (I'd like to say that after I read your comment I also tried your method, too.) So thanks for your advice, but I what I really need is real help. –  Sep 15 '13 at 17:23
  • With a diagonal matrix, it is straightforward to prove that the multiplicity of an eigenvalue is exactly equal to the dimension of the eigenspace. – copper.hat Sep 15 '13 at 17:51
  • Actually, I'm not exactly sure what you are asking. My answer below shows that the algebraic and geometric multiplicities of an eigenvalue are equal (for symmetric matrices). – copper.hat Sep 15 '13 at 20:50

2 Answers2

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In your CASE 1, actually the eigenvectors are orthogonal, and you can normalize them so they are orthonormal. In your CASE 2, eigenvectors corresponding to different eigenvalues are orthogonal, as it seems you know. But eigenvectors corresponding to the same eigenvalue are not necessarily orthogonal unless you construct them or modify them so that they are. For any eigenvalue $\lambda$ whose eigenspace $\{\mathbf{x} \mid A\mathbf{x}=\lambda \mathbf{x}\}$ has dimension greater than $1$ (which occurs if you have "repeated eigenvalues"), use the Gram-Schmidt procedure to find an orthonormal basis for that eigenspace. The Gram-Schmidt procedure is in any linear algebra book and probably on Wikipedia so I won't retype it here. The Gram-Schmidt procedure can be used on any finite-dimensional inner product space and your eigenspace here is such an inner product space.

You also asked why, roughly speaking, the number of eigenvalues is the same as the number of eigenvectors for a real symmetric matrix. This question has been asked and answered here: geometric multiplicity= algebraic multiplicity for a symmetric matrix The "geometric multiplicity" of an eigenvalue is the dimension of its eigenspace, and the "algebraic multiplicity" is the multiplicity of the eigenvalue as a root of the characteristic polynomial.

Of course, I am only talking about real matrices, and when I use the word "dimension" I am using the field of real numbers.

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Stefan Smith
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We can suppose that $A$ is diagonal, with real entries, which are the eigenvalues of $A$. That is, $A$ has the form $A = \begin{bmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n\end{bmatrix}$, with eigenvalues $\lambda_1,...,\lambda_n$.

Let $I_\lambda = \{ k \in \{ 1,...,n\} | \lambda_k = \lambda \}$ be the indices for eigenvalues equal to $\lambda$. The algebraic multiplicity of eigenvalue $\lambda_k$ is then given by $| I_{\lambda_k} |$.

Then it is straighforward to see that $A v = \lambda v$ (equivalently $v \in \ker (A-\lambda I)$) iff $[v]_k = 0 $ for all $k \notin I_\lambda$.

So, suppose $\lambda$ is an eigenvalue, and let $E = \ker (A- \lambda I)$. Then we want to show that $\dim E = | I_\lambda |$.

We see that $v \in E$ iff $v \in \operatorname{sp} \{e_k\}_{k \in I_\lambda}$, that is $E = \operatorname{sp} \{e_k\}_{k \in I_\lambda}$. It follows from this that $\dim E = | I_\lambda |$.

We note that the $\{e_k\}_{k \in I_\lambda}$ are orthogonal, so obviously we can choose an orthogonal basis for the eigenspace $E$.

copper.hat
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