# Showing that the normalized eigenvector for a distinct eigenvalue is unique

Hey guys,

I've been trying to brush up on my linear algebra and ran into this bit of confusion.

I just went through a proof that an operator with distinct eigenvalues forms a basis of linearly independent eigenvectors.

But the proof relied on a one to one mapping of eigenvalues to eigenvectors. Is there any particular reason why for a distinct eigenvalue, there shouldn't be more than one (normalized) eigenvectors that satisfies the eigenvalue definition.

And if so, how do I prove it? I'm not mentally convinced, even if that is the case as the proofs seem to indicate!

Thanks!

HallsofIvy
Homework Helper
Hey guys,

I've been trying to brush up on my linear algebra and ran into this bit of confusion.

I just went through a proof that an operator with distinct eigenvalues forms a basis of linearly independent eigenvectors.

But the proof relied on a one to one mapping of eigenvalues to eigenvectors. Is there any particular reason why for a distinct eigenvalue, there shouldn't be more than one (normalized) eigenvectors that satisfies the eigenvalue definition.
No, there isn't any reason and unless that proof was dealing with a special situation (such as an n by n matrix having n distinct eigenvalues) it isn't true that there is a "one to one mapping of eigenvalues to eigenvectors". For example the diagonal matrix
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$
has the single eigenvalue 2 but every vector is an eigenvector. Even requiring normalization, every unit vector in every direction is an eigenvector.

Now, if you mean, not just "distinct eigenvalues" but "n distinct eigenvalues for an n by n matrix", yes that is true. It follows from the fact that eigenvectors corresponding to distinct eigenvalues are independent. If matrix A is n by n, it acts on an n dimensional space. If A has n distinct eigenvalues, then it has n independent eigenvectors which form a basis for the space. There is no "room" for any other eigenvectors.

And if so, how do I prove it? I'm not mentally convinced, even if that is the case as the proofs seem to indicate!

Thanks!

AlephZero
Homework Helper
For example the diagonal matrix
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$
has the single eigenvalue 2 but every vector is an eigenvector.

I don't think that is what most linear algebraists would call that a "single" eigenvalue, any more than you would say that the equation ##(x- 2)^3 = 0## has only a "single" root (and of course the two statements are closely related).

I thnk the OP's question is about an eigenvalue with multiplicity one, which is what "distinct" means IMO.

No, there isn't any reason and unless that proof was dealing with a special situation (such as an n by n matrix having n distinct eigenvalues) it isn't true that there is a "one to one mapping of eigenvalues to eigenvectors". For example the diagonal matrix
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$
has the single eigenvalue 2 but every vector is an eigenvector. Even requiring normalization, every unit vector in every direction is an eigenvector.

Now, if you mean, not just "distinct eigenvalues" but "n distinct eigenvalues for an n by n matrix", yes that is true. It follows from the fact that eigenvectors corresponding to distinct eigenvalues are independent. If matrix A is n by n, it acts on an n dimensional space. If A has n distinct eigenvalues, then it has n independent eigenvectors which form a basis for the space. There is no "room" for any other eigenvectors.

Oops. Sorry for the vague language, but when I said distinct eigenvalues I did indeed mean multiplicity of 1!

But anyway, where does that fact follow from?
My understanding of the proof that eigenvectors of distinct eigenvalues are independent is something like this (in the special case of n distinct eigen values)
1) The eigenvectors span the null space
2) There are n eigenvectors since there n distinct eigenvalues
3) Since the n = dim, they must all be independent and form a basis

but step 2 assumes that each eigenvalue produces a single eigenvector