1. ## Prove, about characteristic polynomial...

If $A$ is a square matrix of order $n$, then prove that
$det(\lambda I-A)=\lambda ^n-tr(A)\lambda ^{n-1}+P_{n-2}(\lambda)$
with $P_{n-2}$ is the polynomial of order (teh greatest) $n-2\text{ in }\lambda$

2. Originally Posted by GOKILL
If $A$ is a square matrix of order $n$, then prove that
$det(\lambda I-A)=\lambda ^n-tr(A)\lambda ^{n-1}+P_{n-2}(\lambda)$
with $P_{n-2}$ is the polynomial of order (teh greatest) $n-2\text{ in }\lambda$

Follows at once from the definition of char. polynomial and the definition of matrix determinant: if $A=(a_{ij})$ , then $|\lambda I-A|=\left|\begin{pmatrix}\lambda-a_{11}&\ldots & -a_{1n}\\ \ldots &\ldots &\ldots\\-a_{n1}&\ldots &\lambda-a_{nn}\end{pmatrix}\right|$.

Well, check what the coefficient of $\lambda^{n-1}$ is above: precisely the same as in $(\lambda-a_{11})\cdot\ldots\cdot(\lambda-a_{nn})$ , i.e. $-tr.(A)$ .

Tonio