TITLE: Eigenvector proof for repeated eigenvalues
QUESTION [2 upvotes]: I am stuck trying to solve the following problem:
In diagonalizing a symmetric matrix $S$, we find that two of the eigenvalues ($\lambda_1$ and $\lambda_2$) are equal but the third ($\lambda_3$) is different. Show that any vector which is normal to $\hat{n}_3$ (which is the eigenvector corresponding to $\lambda_3$) is then an eigenvector of $S$ with eigenvalue equal to $\lambda_1$
Can anyone offer hints on, or an outline of, the solution?
Thank you.

REPLY [1 votes]: If $A=A^T$ and $\lambda_1, \lambda_2$ are distinct eigenvalues and $\mathbf x, \mathbf y$ corresponding eigenvectors, we have:
$$
\lambda_1\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda_1\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\lambda_2\mathbf{y}\rangle = \lambda_2\langle\mathbf{x},\mathbf{y}\rangle
\Rightarrow
$$
$$
(\lambda_1-\lambda_2)\langle\mathbf{x},\mathbf{y}\rangle=0
$$
and, since $\lambda_1 \ne \lambda_2$, this means that  $\mathbf x, \mathbf y$ are orthogonal. In other words: 

the eigenspaces corresponding to different eigenvalues are orthogonal.

Since $A$ is diagonalizable then we can chose orthogonal eigenvectors in each eigenspace such that all these vectors are an orthogonal basis for $\mathbb{R}^n$.  And this means that a vector orthogonal to an eigenspace is an eigenvector of some other eigenvalue.
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