2012-01-03

A relationship between eigenvalues and determinant


This year started with heartbreak. The multiplication of all the  eigenvalues is determinant of the matrix. (and the sum of all the  eigenvalues is trace of the matrix. These properties are amazing!)  I read an astonishing simple proof using diagonalization of a matrix. This blog is about that proof.

Matrix A can be diagonalize to
Actually, the way to reach this is a bit long and the proof I will show you here is based on this. But, I hope you could agree this. The determinant of this relationship is

By the way,

Because,

A determinant is a scalar value, therefore, they are commutative. Therefore,
What a simple proof this is! But, one thing we need care: this is only true when $S^{-1}$ exists. So, this is not a complete proof, but, I like this simple proof.

References

  • Gilbert Strang, Introduction to Linear Algebra, Chapter 6.

2 comments:

Pedro Peixoto said...

This is only true is A is diagonalizable (for example if it is symmetric).

Shitohichi said...

Hi Pedro,
Yes, you are right. Although, I have already written it in the text: this is only true when $S^{-1}$ exists.
If it doesn't exist, the matrix can not be diagonalizable.
Thanks for the comment!