if we find the eigen values for Matrix A $$\begin{pmatrix}0 & 0 & -2\\1 & 2 & 1 \\1 & 0 & 3\end{pmatrix}$$
I got 2 and 1 with 2 repeat twice
And I Got eigen vectors $$\begin{pmatrix} 1 & 0 & -1 \end{pmatrix}$$
$$\begin{pmatrix} 0&1&0\end{pmatrix}$$
for 2 and
$$\begin{pmatrix}-2&1&1\end{pmatrix}$$ for 1
I want to find P such that A= $P^{-1}$.D.P
where D is diagonal matrix
I know that columns in P are eigen vectors corresponding to eigen value in D. if my D= $$\begin{pmatrix} 2 & 0 & 0 \\\ 0 & 2 & 0 \\\ 0 & 0 & 1 \end{pmatrix}$$
which vector coming first since there are two vectors for two How to determine first vector situation like that(not only for this problem)?
If P=$$\begin {pmatrix} 1 &0&-2 \\\ 0&1&1 \\\ -1 &0&1\end{pmatrix}$$i got answer but when P=$$\begin {pmatrix} 0&1&-2 \\\ 1&0&1 \\\ 0&-1&1\end{pmatrix}$$i didn’t get back A correctly how to determine which vector coming first since (1 0 -1),(0 1 0) both are coming from eigen value 2
Your matrix is correct
For each vector:
$Av = \lambda v$
So for the set:
$AP = PD$
$PDP^{-1} = A$
or $D = P^{-1}AP$
You should have:
$\begin{bmatrix} 0&1&-2\\1&0&1\\0&-1&1\end {bmatrix}\begin{bmatrix} 2\\&2\\&&1\end{bmatrix}\begin{bmatrix} 1&1&1\\-1&0&-2\\-1&0&-1\end {bmatrix} = A$
$\begin{bmatrix} 1&0&-2\\0&1&1\\-1&0&1\end {bmatrix}\begin{bmatrix} 2\\&2\\&&1\end{bmatrix}\begin{bmatrix} -1&0&-2\\1&1&1\\-1&0&-1\end {bmatrix} = A$