How do "Eigenbases" and "orthonormal bases" relate?

Question

So I have a definition for each of the above:

Eigenbasis: when the matrix in question is in diagonal form. Only possible when there are n eigenvectors for a matrix in n-dimensional space.

Orthonormal basis: when basis vectors are a)orthogonal, b)unit-length.

What I'm trying to understand is - how do they relate? Like, is eigenbasis always orthonormal? Or does orthonormal basis always have an eigenbasis? Or are the two the same even?

Thanks!

Answer 1

The eigenvectors of an eigenbasis can be made unitary of course but they are not necessarily mutually orthogonal. This happens, or may happen, when the matrix is hermitian (that is symmetric for real case).

Refer to Spectral theorem.

Answer 2

An eigenbasis need not be orthogonal if the matrix has one or more degenerate eigenvalues. As basis elements need not be unit vectors, an orthogonal basis need not be orthonormal. Any basis is an eigenbasis of a suitably chosen matrix.

Answer 3

Do talk of an eigenbasis you need to refer to some matrix, so the question "Or does orthonormal basis always have an eigenbasis?" is ill-posed.

Are eigenbases always orthonormal?

They are not. A matrix $A$ has an orthonormal eigenbasis if and only if it is normal.

Still, even if $A$ is normal, this doesn't necessarily imply that any eigenbasis of $A$ is orthogonal (although you can always find one such eigenbasis). Indeed, if $A$ has degenerate eigenvalues (i.e. not all eigenvalues are different), then some eigenspaces are more than one-dimensional, and can be decomposed using arbitrary, possibly non-orthogonal, bases.

E.g. if $A=I$ is the $2\times2$ identity, then any pair of linearly independent vectors is an eigenbasis for the underlying space, meaning that there are eigenbases that are not orthonormal. On the other hand, it is trivial to find eigenbases that are orthonormal (namely, any pair of orthogonal normalised vectors).

Finally, it's worth noting that eigenbases need not be made up of normalised vectors. Coming back to the $A=I_2$ example above, $$\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}1\\1\end{pmatrix}\right\}$$ is a perfectly good eigenbasis which is composed of non-normalised, non-orthogonal vectors.

See also the answers to this question for more details.