# Sum and product of eigenvalues

• Dec 7th 2010, 10:36 AM
Sum and product of eigenvalues
How to prove that
- the sum of the eigenvalues of a matrix equals to the trace of the matrix?
- the product of the eigenvalues equals to the determinant?
(Of course the matrix is a square matrix)

I would be very grateful if you could help me.
• Dec 7th 2010, 11:07 AM
FernandoRevilla
Consider $A\in{\mathbb{C}^{n\times n}}$ then, $A$ is similar to its canonical form of Jordan $J\in{\mathbb{C}^{n\times n}}$.

Now use that similar matrices have the same trace and the same determinant.

Regards.

Fernando Revilla
• Dec 7th 2010, 03:13 PM
Drexel28
Quote:

How to prove that
- the sum of the eigenvalues of a matrix equals to the trace of the matrix?
- the product of the eigenvalues equals to the determinant?
(Of course the matrix is a square matrix)

I would be very grateful if you could help me.

Or you could notice that if $A=[a_{i,j}]$ the the characteristic polynomial is given by

$\displaystyle p_A(x)=\prod_{j=1}^{n}(x-\lambda_j)$

Thus, recalling Vieta's formulas and noticing that the characteristic polynomial is monic we can see the the constant term of the above is $\displaystyle (-1)^n\prod_{j=1}^{n}\lambda_j$. That said, we know that for any polynomial $q(x)$ the constant term is merely $q(0)$. Thus, with equal validity we may conclude that the constant term of $p_A(x)$ is

$p_A(0)=\det\left(I0-A\right)=\det\left(-A\right)=(-1)^n\det\left(A\right)$

comparing these two gives $\displaystyle \det(A)=\prod_{j=1}^{n}\lambda_j$ as you required. A similar technique works for the trace.

Remark: Vieta's formula (for the above, not the trace proof) is a bit of an overkill. It's easy to see that what I claimed was the constant term is, in fact, the constant term.