I tried to answer the following question but stuck in number three.
\begin{pmatrix}
-1 & 1 & 0 \\
0 & 2 & 2\\
0 & 1& 1
\end{pmatrix}
1. find the eigenvalue
$\lambda=3, \lambda=-1,\lambda=0$
2. find eigenvector
when $\lambda=0$
$v_1=
\begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}$
when $\lambda=3$
$v_2=
\begin{pmatrix}
\frac{1}{2} \\
2 \\
1
\end{pmatrix}$
when $\lambda=-1$
$v_3=
\begin{pmatrix}
1 \\
0 \\
0
\end{pmatrix}$
3. find matrix $P$ that diagonalize $A$, $P^{-1}AP$
here I was confused, although I got all three eigen vector and geometric multiplicity for all eigen value same as algebraic multiplicity but the matrix $A$ itself has $\lambda=0$ as one of its eigen value which mean that the matrix $A$ dont have inverse, and $\det |A|=0$, rank $A$ is $2$, which mean $A$ is linear dependent. So is it mean that there isn't matrix $P$ that can diagonalize it? Is there any relationship between eigen value $\lambda=0$ and invertible and diagonalizable?
and is my approach true?
if this true should i use gram schmidt to find the ortogonal matrix to diagonalize it? or should i just find the inverse of P? thanks!!
Indeed, the matrix $A$ has no inverse. Nobody claimed it has. Just take$$P=\begin{pmatrix}1&\frac12&1\\1&2&0\\-1&1&0\end{pmatrix}.$$Note that the columns of $P$ are your eigenvectors (disclaimer: I did not check your computations). Then$$P^{-1}AP=\begin{pmatrix}0&0&0\\0&3&0\\0&0&-1\end{pmatrix}.$$