Are orthonormal eigenbases unique?

[Bipolarity]

Suppose you have an operator $T: V → V$ on a finite-dimensional inner product space, and suppose it is orthogonally diagonalizable. Then there exists an orthonormal eigenbasis for V. Is this eigenbasis unique?

Obviously, in the case of simple diagonalization, the basis is not unique since scaling (by nonzero) any vector in an eigenbasis yields a valid eigenbasis.

Likewise, an orthonormal basis for a space of at least dimension 2 is not unique, since we can take any two nonparallel vectors in the space and extend each to its own orthonormal basis through Gram-Schmidt. The two bases must be distinct.

But what about an orthonormal eigenbasis? Is this set unique? My guess is that it is, but I need to know for sure so I can think about which direction I want to steer my proof.

Thanks!

BiP

 

[jgens]

Psst...consider the identity map and two distinct orthonormal bases for V.

Edit: There are plenty of other examples too. This one is just the easiest.

 

[Office_Shredder]

You can decompose your vector space V uniquely into
$$V = \bigoplus_{\lambda} V_{\lambda}$$
where $V_{\lambda}$ is the eigenspace of eigenvalue $\lambda$. Inside of each eigenspace if the dimension is not equal to 1 you have a lot of freedom as to what basis you pick (as jgens example gives). Even if each eigenspace is 1 dimensional, you still have two unit vectors from each eigenspace that you can pick, so there are still 2ⁿ orthornomal eigenbases.

 

[AlephZero]

Even if each eigenspace is 1 dimensional, you still have two unit vectors from each eigenspace that you can pick, so there are still 2ⁿ orthornomal eigenbases.

Unless I'm missing the point completely, I suppose you mean vectors V and -V?

If you count those as "different" vectors, then if V is complex there are infinite number of vectors to pick from. You can multiply the vector by $e^{i\theta}$ for any $\theta$.

FWIW in practical engineering eigenvalue problems, you often make a further arbitrary choice about normalization (e.g. make the element of the vector with the maximum modulus real and positive) to nail down those arbitrary but annoying differences, when comparing results from slightly different eigensolutions.

 

[Office_Shredder]

Unless I'm missing the point completely, I suppose you mean vectors V and -V?

If you count those as "different" vectors, then if V is complex there are infinite number of vectors to pick from. You can multiply the vector by $e^{i\theta}$ for any $\theta$.

Oh yeah I forgot sometimes the field isn't $\mathbb{R}$

 

[Bipolarity]

Ahh I forgot that scaling a vector by a number on the unit disk in the complex plane (or -1 if the field is real)) preserves normalization of the vector.

Thank you guys!

BiP