In mathematica, one enters matrices row - by - row as a ' nested list'

A={{-1,-2, -2},{1,2,1},{-1,-1,0}}

If you' d actually like to see what you' ve done, it helps to display it like a matrix

MatrixForm[A]

Anyhow, then you can compute the characteristic polynomial of your matrix and factor it

charpoly = Det[A-t DiagonalMatrix[{1,1,1}]]
Factor[charpoly]

Or you can just ask for the eigenvalues

Eigenvalues[A]

Given the eigenvalues, you can then compute eigenvectors

Solve[A.{x1,x2,x3}== -{x1,x2,x3},{x1,x2,x3}]

Solve[A.{x1,x2,x3}== {x1,x2,x3},{x1,x2,x3}]

Or, as before, you can just ask and receive

Eigenvectors[A]

Note that the eigenvectors are given in the same order as the eigenvalues.  Thus we have
A = Chg Diag Chg^{-1}, where

Diag = DiagonalMatrix[Eigenvalues[A]]

Chg = Transpose[Eigenvectors[A]]

MatrixForm[Diag]
MatrixForm[Chg]

The transpose is necessary to make the eigenvectors columns, rather than rows, of Chg.  Just to be sure, I check my results.

MaybeA = Chg.Diag.Inverse[Chg]

A-MaybeA

Voila