# Homework Help: Linear Algebra Proof Help

1. May 5, 2008

### ltrane2003

1. The problem statement, all variables and given/known data
Prove that the product of the diagonal entries of an nxn matrix A equals the product of the eigenvalues of A.

2. Relevant equations

3. The attempt at a solution

2. May 5, 2008

### AstroRoyale

You have to show at least some attempt first before anyone will help.

3. May 5, 2008

### DavidWhitbeck

Since you haven't tried the problem I won't provide much help, but I will say a couple of vague hints

(a) this is a standard textbook result, see if it's in your book first
(b) examine the coefficients of the characteristic polynomial carefully

4. May 5, 2008

### HallsofIvy

The problem with you not showing any work is that we have no idea what kind of facts you have to work with. I can think of several different ways of proving that, depending upon how "sophiticated" you want to be.

The very quickest would involve using the "Jordan Normal Form"- do you know what that is?

5. May 5, 2008

### Dick

But it's not even true!? Take A=[[0,1],[1,0]]. Product of the eigenvalues is -1. Product of the diagonal entries is 0. Am I missing something?????

6. May 5, 2008

### rock.freak667

That is not true for all matrices...only in certain types of matrices is that true.

7. May 5, 2008

### AstroRoyale

The product of the eigenvalues is the determinant, right. The statement would be true for a diagonal matrix for sure :)

8. May 5, 2008

### HallsofIvy

Yes, it is true (and is trivial) for every diagonal matrix. But that is a very small subset of all matrices.

9. May 5, 2008

### Vid

Both the eigenvalues are zero.

For a diagonal matrix the determinant is just the product of the diagonals.

The eiganvalues are |xI-A|....

10. May 5, 2008

### Dick

A*[1,1]=[1,1]. That doesn't look like a zero eigenvector to me.

11. May 5, 2008

### Vid

Diagonal matrices have 0 everywhere but the diagonal. Your matrix is symmetric.

12. May 6, 2008

### DavidWhitbeck

Yeah I totally screwed up, the OP never returned but I'll just say what I was thinking-- the constant term in the characteristic polynomial is known to be the determinant of the matrix, call it A, but also if you factor the polynomial knowing that it's roots are the eigenvalues $$\lambda_i$$ then the constant term is also

$$(-1)^N\prod_{i=1}^{N}\lambda_i$$

and there you have it--

$$\prod_{i=1}^{N}\lambda_i = (-1)^N\det A$$

where A is an N by N matrix.

13. May 6, 2008

### Dick

Now you have to explain away that (-1)^N. It shouldn't be there. If I is the NxN identity. det(I)=1 and the product of the eigenvalues is certainly 1.

14. May 6, 2008

### DavidWhitbeck

Ah yeah the constant term in the characteristic polynomial is not $$\det A$$ it's $$(-1)^N\det A$$ and then we have instead

$$\prod_{i=1}^{N}\lambda_i = \det A$$