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.

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- Dec 7th 2010, 09:36 AMzadirSum 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, 10:07 AMFernandoRevilla
Consider $\displaystyle A\in{\mathbb{C}^{n\times n}}$ then, $\displaystyle A$ is similar to its canonical form of Jordan $\displaystyle J\in{\mathbb{C}^{n\times n}}$.

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

Regards.

Fernando Revilla - Dec 7th 2010, 02:13 PMDrexel28
Or you could notice that if $\displaystyle A=[a_{i,j}]$ the the characteristic polynomial is given by

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

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

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

comparing these two gives $\displaystyle \displaystyle \det(A)=\prod_{j=1}^{n}\lambda_j$ 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.