# Eigenvalues are unique?

I'm studying eigenvector and eigenvalue but there are some confusing things to me.

(1) Eigenvectors are not unique
(2) If eigenvectors come from distinct eigenvalues, then eigenvectors are unique.

Here is my question. Then, eigenvalues are unique all the time?
Or there are any restrictions to make eigenvalues to be unique just like the case of eigenvectors?

Also, when we diagonalize matrix A=$S\lambda S^{-1}$, the eigenvector matrix S is not unique. It means if we multiply each column of S by nonzero constant we can make a new $S'$.
Why is that?
If we can diagonalize matrix A, it means that eigenvectors are unique. But the eigenvetor matrix S is not unique? Then with the new eigenvector matrix$S'$, $S'\lambda S'^{-1}$ makes the same matrix A?

My last question is that. If eigenvalues are not unique, then eigenvalue matrix $\lambda$ is not unique too? But $S\lambda'S^{-1}$ makes same matrix A?

• What do you mean by "eigenvectors are unique"? Your statement in (2) should end with "eigenvectors are linearly independent". – wj32 Nov 12 '12 at 3:03
• Yes, if the eigenvectors come from distinct eigenvalues, they are linearly independent. It is not corresponding to "unique"? I just searched and found that comment on the internet. – corly Nov 12 '12 at 3:07
• Internet is full with all kinds of nonsenses, from religious, political, and otherwise propaganda all the way down to "proofs" that Cantor's Theorem is wrong or that Washington is the capital of Uzbekistan. You should be careful with this. The phrase "eigenvectors are(not) unique" must make sense for a mathematician to use it, and here it is not clear at all. But trying to make some sense of it, take into account that if $\,u\,$ is an eigenvector of some matrix then so is $\,au\,$ for any scalar $\,a\,$, and this is is far from some interpretation of "unique"... – DonAntonio Nov 12 '12 at 3:13
• Yes. Your language is somewhat vague, and you need to be careful with how you use the word "unique". – wj32 Nov 12 '12 at 3:15
• Then the statement (1) is still right? Eigenvectors are not unique, since any vector can be an eigenvector of the identity matrix. My main question is that, "eigenvalues are unique?" – corly Nov 12 '12 at 3:18

## 2 Answers

There seems to be several points of confusion here, so let me ignore your questions for now and start from the top.

The eigenvalues of a matrix $A$ are defined as the set of values $\lambda$ for which the matrix $A-\lambda I$ is singular. Put another way, the eigenvalues of the matrix $A$ are the set of values $\lambda$ for which $$p(\lambda) = \det(A-\lambda I) = 0$$ The expression $\det(A-\lambda I)$ is called the characteristic polynomial of $A$ and the eigenvalues are defined to be the roots of this polynomial. In general, the characteristic polynomial of an $n\times n$ matrix is an $n$th degree polynomial which means that there will be (at most) $n$ roots of the polynomial. The set of eigenvalues is what we call the spectrum of $A$. The spectrum is the set of values which appears on the diagonal of your diagonal matrix. These values are unique but only up to order.

So let me now address your question: "Are the eigenvalues of a matrix unique?" Well that's a bit difficult to answer because the question is not formulated well. If a matrix has $n$ distinct eigenvalues, would you consider each eigenvalue to be unique (in the sense of multiplicity one)? If a matrix has only a single eigenvalue of multiplicity $n$, would you consider that to be unique?

In either case, the answer to your question would be no. A matrix does not necessarily have distinct eigenvalues (although almost all do), and a matrix does not necessarily have a single eigenvalue with multipicity $n$. In fact, given any set of $n$ values, you can construct a matrix with those values as eigenvalues (indeed just take the corresponding diagonal matrix).

Now onto eigenvectors. For each eigenvalue $\lambda$, there exists a subspace of vectors $E_\lambda$ which satisfies the equation $$A\mathbb{v} = \lambda\mathbb{v}$$ for $\mathbb{v}\in E_\lambda$. Now this eigenspace $E_\lambda$ is unique, but the vectors in the space, the eigenvectors are not unique. It is analogous to the fact that you can talk about there being a unique $x$-axis, but it makes no sense to talk about a unique point on the $x$-axis.

What is true is that the eigenspaces of different eigenvalues are independent, so that eigenvectors of different eigenvalues are linearly independent. When your matrix is diagonalizable, the collection (or direct sum if you are familiar with the term) of these eigenspaces is your entire vector space. This means that there exists a basis of solely eigenvectors and your matrix $S$ is formed from a basis of eigenvectors as its columns. Of course, each eigenspace is in fact a subspace, so linear combinations of eigenvectors remain eigenectors. This is why you can multiply your eigenvectors by scalar multiples and still have them remain eigenvectors. In fact, you are free to choose any basis for the eigenspace and your matrix $S$ will correspondingly be modified.

• I was reading your answer and a doubt arise, we generally say that if a linear system has a singular coefficient matrix then we can either have infinite solutions or no solution. Solving $(A-\lambda I)v=0$ and given that $det(A-\lambda I)=0$ we will have infinite solutions (Am I wrong at this point?) so why do we say that there's one eigenvector for each eigenvalue (since we don't have infinite eigenvalues)? – RScrlli Mar 15 at 7:35

1.Given a matrix, the superset (a set that allows multiple instances of an element) of eigenvalues is unique. It implies that you can not find a different superset of eigenvalues for a matrix.

2.Eigenvectors corresponding to eigenvalues of single multiplicity are parametrized by a coefficient which is denoted by $c.\mathbf{x}$. The unit eigenvector $\mathbf{x}$ is unique up to sign (it can be multiplied by -1) for this case.

3.For eigenvalues of multiplicity $n$, the dimension of eigenspace $m$ is such that $m \leq n$. For this case, we can find different sets of unit vectors which span the same eigenspace. In other words, set of eigenvectors is not unique for this case. (Here, we assume that $m>1$, i.e if $m=1$, the unit eigenvector is unique up to sign as noted in 2.