Finding the characteristic polynomial of this specific $3\times3$ matrix How can I find the characteristic polynomial of the following matrix:
\begin{pmatrix} 0&-2&2\\-2&1&0\\2&0&-1 \end{pmatrix}
please I need the details.
 A: The characteristic polynomial is given by $$ \chi_A(T) = \det(T\cdot E_n - A) $$
where $A$ is your matrix, $E_n$ is the $n$-dimensional unity matrix.
To find it for your specific matrix, you simply replace $A$ by your matrix.
A: The characteristic polynomial of a matrix $A$ is $p_A(\lambda) = \det(A-\lambda I)$. So in your case, you need to expand the expression/determinant $$\begin{vmatrix} 0-\lambda&-2&2\\ -2 &1-\lambda& 0 \\ 2 & 0&-1-\lambda\end{vmatrix}$$ I choose to expand along the row/column with the most number of zero's, but I'll expand across the top row below for the sake of evaluating an expression that resembles a cross product: $$ -\lambda\begin{vmatrix}1-\lambda & 0 \\ 0 & -1-\lambda \end{vmatrix} - (-2)\begin{vmatrix} -2 & 0 \\ 2 & -1-\lambda \end{vmatrix} + 2 \begin{vmatrix} -2 & 1-\lambda \\ 2 & 0 \end{vmatrix} \\ = -\lambda[(1-\lambda)(-1-\lambda)-(0)(0)]+2[(-2)(-1-\lambda)- (0)(2)] + 2[(-2)(0)-(1-\lambda)(2)] \\ = -\lambda[-1-\lambda+\lambda+\lambda^2-0]+2[2+2\lambda-0]+2[0-2+2\lambda] \\ = -\lambda[\lambda^2-1]+2[2\lambda+2]+2[2\lambda-2] \\ = -\lambda^3+\lambda +4\lambda+4+4\lambda-4 \\ = -\lambda^3+9\lambda$$
Therefore, $$p_A(\lambda) = -\lambda^3+9\lambda$$
A: if we have matrix  
\begin{pmatrix} 0&-2&2\\-2&1&0\\2&0&-1 \end{pmatrix}
please substract some $\alpha$ from diagonal terms and  find  determinant,like if we have
 a=[2 1;3 4]

a =

     2     1
     3     4

characteristic polynomial is simple determinant of following matrix
2-k     1 
  3       4-k

A: For $3x3$ matrices use following formula, it's extremely fast once you know your way through it:
$$ \chi_A(T) = \lambda ^3-tr(T)\lambda^2+c_2\lambda-det(T)$$
Where $c_2$ is the sum of the principal minors of the matrix, hence:
$$c_2=(t_{11}t_{22}-t_{12}t_{21})+(t_{11}t_{33}-t_{13}t_{31})+(t_{22}t_{33}-t_{23}t_{32})$$
I've always felt more confortable with this formula because you don't have to change signs nor evaluate multiplications with more terms which may lead to easy mistakes
Remark: Remember that the the coefficient of the term with the second highest monom ($x^{n-1}$) is always the trace and the coefficient of the last one (the constant) is the determinant of the matrix. You can check this properties quite trivially. Pay attention to the fact that I use the sign convention with $+$ for the first term.
PS: such a formula exists also for $2 \times 2$ and $4 \times 4$ matrices, you might want to check them out here: http://en.wikipedia.org/wiki/Characteristic_polynomial .
