In an inner product space, if the matrix is symmetric, is an eigenspace necessarily orthogonal to the range space?

Question

Say I have 3 distinct eigenvalues for a symmetric matrix.

By the Spectral Theorem, the three eigenspaces are mutually orthogonal.

But, if I just wanted to compute the first eigenspace, corresponding to $\lambda_1$, would this eigenspace, which is the null space of the matrix A-$\lambda_1$, be orthogonal to the range space of A-$\lambda_1$?

I feel that all we can really say is that the nullity and rank of A-$\lambda_1$ add up to give the dimension of the whole space, and that there isn't an orthogonality connection here, except for when we compare eigenspaces with eigenspaces.

Thanks,

Answer 1

Hint: The null space of $A$ is always orthogonal to the range of $A^T$, for any matrix $A$. Now, exploit that $A$ is symmetric.

Solution:

Let $x$ be in the null space of $A$ and $y=A^T u$ be in the range of $A^T$. Then, we have $$ x^T y = x^T A^T u = (Ax)^T u = 0^T u = 0. $$ Since $x$ and $y$ were arbitrary, it follows that the null space of $A$ and the range of $A^T$ are orthogonal.

If $A$ is symmetric, then the dimension formula shows that the null space of $A$ is in fact the orthogonal complement of the range of $A$.

Answer 2

Yes, eigenvectors corresponding to different eigenvalues are necessarily independent so, in an inner-product space, orthogonal.

If an n by n matrix is symmetric then there are $n$ independent eigenvectors. Therefore, there exist eigenvectors, each of which spans a one dimensional vector space so the eigenvectors form an orthogonal basis for the entire space.