# If a real symmetric matrix has repeated eigenvalues, why does it still have n linearly independent eigenvectors?

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)?

• 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. – levap Nov 7 '15 at 16:36
• 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 $b\neq c$ – janmarqz Nov 7 '15 at 16:41
• @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? – Sibbs Gambling Jul 25 '16 at 3:47
• @SibbsGambling: Yep, that is correct. – levap Jul 25 '16 at 3:49
• Consider the case $A = 1$ for comparison. – anomaly Aug 23 '18 at 19:47

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 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.