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

j3dwards
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Homework Statement


Explain in your own words why the product of eigenvalues of any diagonalisable N × N matrix A must equal the determinant of A.

Homework Equations


MT=M-1

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?
 
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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.
 
In the title and in your last sentence it should be "eigenvalues" instead of eigenvectors.
j3dwards said:
##M^T=M^{-1}##
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.)
 
blue_leaf77 said:
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.

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

Please explain, I really don't understand and my exam is soon!
 
j3dwards said:
You get the eigenvalues in the diagonal elements in the diagonalised matrix.
Yes, and if you calculate the determinant of this diagonal matrix, how does it look like in terms of the eigenvalues?
 
j3dwards said:
You get the eigenvalues in the diagonal elements in the diagonalised matrix.

Please explain, I really don't understand and my exam is soon!
What happens if you take the determinant of both sides of ##D = P^{-1}AP## ?
 

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