1. ## Eigenvalue proof.

Prove that every singular matrix A has 0 as an eigenvalue.

I don't really have a clue how to prove this.

I can see that when A is a singular matrix, the eigenvalues are found using the equation A lamda^2 + B lamda = 0 so one value of lamda is always 0.

I don't know how to prove it though..

2. Originally Posted by chr91
Prove that every singular matrix A has 0 as an eigenvalue.

I don't really have a clue how to prove this.

I can see that when A is a singular matrix, the eigenvalues are found using the equation A lamda^2 + B lamda = 0 so one value of lamda is always 0.

I don't know how to prove it though..
There are multiple results which finish this problem immediately, so it depends what you know. Probably the easiest method is to note that if $\displaystyle \sigma\left(A\right)$ denotes the multiset of eigenvalues (i.e. all the eigevalues, including multiples) then $\displaystyle \displaystyle \det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda$. Thus, if $\displaystyle A$ is singular then $\displaystyle ]\displaystyle 0=\det A=\prod_{\lambda\in\sigma\left(A\right)}\lambda$ from where it follows that $\displaystyle \lambda$ is in whatever the underlying field you're working with (if that last part didn't make sense replace it with "since the $\displaystyle \lambda$s are in $\displaystyle \mathbb{C}$) it follows that at least one element of $\displaystyle \sigma\left(A\right)$ must be zero.

Anothe direction is that $\displaystyle A$ is singular if and only if $\displaystyle Ax=\mathbf{0}$ for some $\displaystyle x\ne \mathbf{0}$, but I doubt you know what since this is, basically the question.

3. $\displaystyle \displaystyle det(A)=(a_{11}-\lambda)A_{11}+\sum_{i=2}^{n}a_{i1}A_{i1}$

$\displaystyle \displaystyle (a_{11}-\lambda)A_{11}=(a_{11}-\lambda)(a_{22}-\lambda)...(a_{nn}-\lambda)$

$\displaystyle \displaystyle =(-1)^n\lambda^n+...+(-1)^{n-1}\lambda^{n-1}$

$\displaystyle \displaystyle p(0)=det(A)=\lambda_1\lambda_2...\lambda_n$

$\displaystyle \displaystyle (-1)^{n-1}=tr(A)=\sum_{i=1}^{n}\lambda_i$

$\displaystyle \displaystyle p(\lambda)=0$ has exactly n solutions $\displaystyle \displaystyle \lambda_1,...,\lambda_n$

$\displaystyle \displaystyle p(\lambda)=(\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)$

$\displaystyle \displaystyle p(0)=(\lambda_1)(\lambda_2)...(\lambda_n)=det(A)$

4. Originally Posted by chr91
Prove that every singular matrix A has 0 as an eigenvalue.
An alternative:

$\displaystyle A\;\textrm{singular}\;\Rightarrow \det A=0 \Rightarrow \det (A-0I)=0\Rightarrow \lambda=0\textrm{\;is\;eingenvalue\;of\;}A$

Regards.

Fernando Revilla