# Why is the product of eigenvalues equal to the det(A)?

1. Jan 14, 2016

### j3dwards

1. The problem statement, all variables and given/known data
Explain in your own words why the product of eigenvalues of any diagonalisable N × N matrix A must equal the determinant of A.

2. Relevant equations
MT=M-1

3. The attempt at a solution
So what I do know: the determinant measures the change in area of the unit square under the transformation (as the point (x,y) transforms to the point (X,Y)). And the eigenvectors describe the direction of the deformation of the matrix A - which are unchanged by the deformation.

So my question is why does the product of the eigenvectors equal the determinant of the matrix A?

2. Jan 14, 2016

### blue_leaf77

If you diagonalize a diagonalizable matrix, what do you get as the diagonal elements in the diagonalized matrix? Another hint, determinant of a matrix is invariant under change of basis.

3. Jan 14, 2016

### Krylov

In the title and in your last sentence it should be "eigenvalues" instead of eigenvectors.
Why are you assuming $M$ is unitary? This is not necessary.

Hint: Jordan normal form.
Edit: Start with @blue_leaf77's suggestion, then use the Jordan form for the general case. (From the title I got you had to solve it in general, from your description I get you may assume that $M$ is diagonalizable.)

4. Jan 14, 2016

### j3dwards

You get the eigenvalues in the diagonal elements in the diagonalised matrix.

Please explain, I really don't understand and my exam is soon!

5. Jan 14, 2016

### blue_leaf77

Yes, and if you calculate the determinant of this diagonal matrix, how does it look like in terms of the eigenvalues?

6. Jan 14, 2016

### LCKurtz

What happens if you take the determinant of both sides of $D = P^{-1}AP$ ?