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

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

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

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

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

Tonio