Show that the matrix is diagonalizable and find a diagonal matrix similar to the given matrix.
$\displaystyle
\begin{pmatrix}0 & -2 & 1\\1 & 3 & -1\\0 & 0 & 1\end{pmatrix}
$
Use this same proceedure to find the matrix P
http://www.mathhelpforum.com/math-he...ar-matrix.html
actually you don't need the matrix P. You will find that the char poly has a repeated eigenvalue of $\displaystyle \lambda =1$
You will need to show that $\displaystyle \lambda =1$ gives two eigen vectors.
After you have shown this the similar matrix will be the matrix with the eigenvalues of the char. poly down the diagonal.
I got 2, 1, 1 as the eigenvalues from the characteristic polynomial $\displaystyle x^{3}-4x^{2}+5x-2$. Do the numbers of the diagonal follow any order? Would it be $\displaystyle \begin{pmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 1\end{pmatrix}$ or $\displaystyle \begin{pmatrix}2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ Thank you!
Your eigenvalues and vectors are correct .
What I am saying is this
If you make your matrix P such that $\displaystyle PAP^{-1}=D$
like this this
$\displaystyle \begin{bmatrix}
-1 & 1 & -2 \\
1 & 0 & 1 \\
0 & 1 & 0 \\
\end{bmatrix}
$
Where the first column is the vector corrisponding to the eigenvalue 2 then the diagonal matrix will look like this
$\displaystyle \begin{bmatrix}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$
If you move exchange the first two columns P would look like this
$\displaystyle \begin{bmatrix}
1 & -1 & -2 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
\end{bmatrix}
$
Then D would look like this
$\displaystyle \begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$
Both of the above are similar to the original matrix.
I hope this clears it up.