# Eigenvalues redux

1. Mar 18, 2005

### gnome

matt and cronxeh seemed to imply that we should all know that the product of the eigenvalues of a matrix equals its determinant. I don't remember hearing that very useful fact when I took linear algebra (except in the case of diagonal matrices, where it's obvious), and I can't find it mentioned in the textbook. Sure enough, it's true for every matrix that I've tried, but I can't see how to derive it. Would one of you please show how?

2. Mar 18, 2005

### Tom Mattson

Staff Emeritus
When you diagonalize a matrix (that is, when you write it in the basis of its own eigenvectors) the eigenvalues appear on the diagonal. That, combined with the fact that the trace of a matrix is independent of basis, proves the point.

3. Mar 18, 2005

### shmoe

Alas, not all matrices are diagonalizable. In the general case, take a look at the characteristic polynomial, especially it's constant term.

4. Mar 18, 2005

### hypermorphism

The determinant of a transformation is the oriented n-volume of the parallelotope formed by the transformed basis relative to the old basis.
Ie., let V be R^3 with the standard basis, and let T be a linear operator on V that doubles the magnitude of all vectors. The T has the matrix representation (2,0,0;0,2,0;0,0,2). By the above definition, equivalent to the algebraic formula, the determinant is then 8. Note that inverting any one of the axes causes the oriented 3-volume of the parallelepiped to be negative, because the orientation of the axes has changed (you cannot rotate the old basis into the new basis if the new basis has inverted an odd number of axes). Zero-ing any one of the basis vectors causes the 3-volume to go to 0, as we do not want to confuse 2-volume (area) with 3-volume.
It is then immediately apparent that if the transformed basis is just a rescaling of the old basis, then the oriented n-volume of the new basis with respect to the old basis is just the product of the scalars (the eigenvalues). Depending on the way you learned linear algebra, this geometric motivation for the determinant may not have been made apparent.

Last edited: Mar 18, 2005
5. Mar 18, 2005

### gnome

Alas, indeed. I've been reading and re-reading the theorems about which matrices are diagonalizable, and I'm more or less comfortable with them. But it appears that the answer to my question is going to be much more complex than I expected.

For example, this cryptic statement

6. Mar 18, 2005

### gnome

That's easy for you to say. :yuck:

Maybe I'll just accept it & worry about the details some other time.

7. Mar 18, 2005

### shmoe

The characteristic polynomial of an nxn matrix is det(xI-A), (or det(A-xI) in some books).

Suppose this thing factors completely into linear factors:

$$det(xI-A)=(x-\lambda_1)\dots (x-\lambda_n)$$

Then the lambdas are the eigenvalues. The constant term on the left can be found by sticking in x=0 to get det(-A). The constant term on the right is just the product of the lambda's times (-1)^n (or stick x=0 into the right side as well). That's your result.

If you're looking at a matrix with real coefficients and it's characteristic polynomial doesn't factor completely in the reals, you can factor it over the complex numbers and get a result if you are willing to allow complex eigenvalues.

If you wanted to beat this thing down with a big hammer, you could look up Jordan Form, or Schur's theorem about matrices being similar to upper triangular ones (doesn't always carry schur's name)

For a bonus lolipop, can you see where the sum of the eigenvalues appers in the characteristic polynomial?

Last edited: Mar 18, 2005
8. Mar 18, 2005

### shmoe

You're assuming here that you have a basis of eigenvectors, so this will only work for diagonalizable matrices. Maybe I'm missing something obvious?

9. Mar 18, 2005

### cronxeh

if A can be diagonalized, then it can be rewritten as $$A = PDP^{-1}$$ where D is diagonal matrix of all eigenvalues of A, and P is the invertible matrix consisting of all eigenvectors of A (which correspond to all eigenvalues in D). Obviously determinant of $$PDP^{-1}$$ will be multiple of all eigenvalues in the diagonal matrix, and the trace will be the sum of all eigenvalues, since $$P*P^{-1} = I$$

$$A$$ can be diagonalized if and only if rank of the matrix formed by eigenvectors of $$A_{nxn}$$ = n. That is, there are n linearly independent eigenvectors of A.

Last edited: Mar 18, 2005
10. Mar 18, 2005

### gnome

Yeah, that part I got. I'm just trying to comprehend the situations where that doesn't apply, which I realize are simply not addressed at all in the texts (or the PARTS of the texts) that I've read.

11. Mar 18, 2005

### hypermorphism

Heya shmoe,
You're right, that was too specific. I should have said "...if the transformed basis is just a rescaling of any basis of V,...", which segues right into the "diagonalizable" business.

Last edited: Mar 18, 2005
12. Mar 18, 2005

### gnome

One of my books refers to the characteristic equation as $\text{det}(A-\lambda I) = 0$ and the other states it as $\text{det}(\lambda I - A) = 0$. I haven't gotten up to dealing with complex eigenvalues yet. So, sticking to your linear factors example:

it's been prettly clear to me that we can find the eigenvalues, assuming it factors nicely, by solving
$$(\lambda-\lambda_1)\dots (\lambda-\lambda_n) = 0$$
and clearly each of those $\lambda_i$ is an eigenvalue. I don't understand what you are doing with your "x=0 to get det(-A)" or "x=0 into the right side ..." leaving the product of all the $\lambda_i$s.

13. Mar 18, 2005

### shmoe

Either definition of the characteristic polynomial is fine, they only differ by (-1)^n.

So you're ok with

$$det(xI-A)=(x-\lambda_1)\dots (x-\lambda_n)$$

Now substitute x=0 in on both sides:

$$det(0I-A)=(0-\lambda_1)\dots (0-\lambda_n)$$

$$det(-A)=(-\lambda_1)\dots (-\lambda_n)$$

(since A is nxn, remember how multiplying by a constant affects it)

$$(-1)^{n}det(A)=(-1)^{n}\lambda_1\dots \lambda_n$$

$$det(A)=\lambda_1\dots \lambda_n$$

14. Mar 18, 2005

### Data

Look up Jordan Normal Form (or equivalently Jordan Canonical Form) if you're really interested in this. The triangular form theorem would also be sufficient for a discussion of determinants.

Essentially, the theorem says that every n x n matrix A over $$\mathbb{C}$$ (this requirement is not part of the theorem as usually stated, of course, and can easily be generalised, but you have to be a little bit careful, since if the minimal polynomial of A is not factorizable into linear factors the triangular form theorem does not apply) is similar to at least one upper triangular (upper triangular is actually weaker than the theorem's statement, but it's okay for this discussion) matrix. Obviously in the special case that A is diagonalizable, then we can just say A is similar to a diagonal matrix.

It follows that since similar matrices have the same characteristic polynomial, that the diagonal entries of such an upper triangular matrix must be A's eigenvalues (each repeated a number of times equal to its algebraic multiplicity). But of course, the product of the diagonal entries is just the determinant of A, which answers your question.

There are quite a few results necessary before the triangular form theorem becomes obvious.

Last edited: Mar 18, 2005
15. Mar 18, 2005

### mathwonk

heres a possible argument: an equation like detA = product of eigenvalues, is true on a closed set. thus since diagonalizable matrices are dense, it is true for all matrcies iff it is true for diagonalizable ones.

16. Mar 18, 2005

### gnome

shmoe, I'm not stuck at the end of your argument, but the beginning. You're starting with $det(xI-A)=(x-\lambda_1)\dots (x-\lambda_n)$ where x is some eigenvalue of A, right?

So, if you set x=0 in that equation to show that $det(A)=\lambda_1\dots \lambda_n$, haven't you proven it only for a matrix that has 0 as an eigenvalue?

edit: forget about this. For some reason I was looking at the derivation of this equation completely backwards, as if we were starting with these eigenvalues $\lamba_1 \dots \lambda_n$ and then ... etc etc

Last edited: Mar 19, 2005
17. Mar 18, 2005

### Data

Why would that be true? It's an equation, like any other. If 0 were an eigenvalue, you would get a factor of $x$ multiplying everything on the right, and would find $$\mbox{det}(A)=0$$.

18. Mar 19, 2005

### gnome

OK, I see that now. Thanks.

19. Mar 19, 2005

### matt grime

The simple product of the eigenvalues is not the determinant - there are mutliplicities to worry about.

The constant in Det(A-xI) corresponds to the product of roots of the characteristic polynomial with multuiplicity assuming certain things about the base field containing enough roots of unity.

Last edited: Mar 19, 2005
20. Mar 19, 2005

### gnome

This:
I think is apparent if the rhs is simply a factorization of the lhs, right?

but I don't know what this:
means, particularly, what is the constant you referring to, and can you state "the base field containing enough roots of unity" in my non-mathematician's English, or should I simply not worry about it?