1. ## Diagonal Matrix

Show that the matrix is diagonalizable and find a diagonal matrix similar to the given matrix.
$
\begin{pmatrix}0 & -2 & 1\\1 & 3 & -1\\0 & 0 & 1\end{pmatrix}
$

2. Originally Posted by jennifer1004
Show that the matrix is diagonalizable and find a diagonal matrix similar to the given matrix.
$
\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 $\lambda =1$

You will need to show that $\lambda =1$ 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.

3. I got 2, 1, 1 as the eigenvalues from the characteristic polynomial $x^{3}-4x^{2}+5x-2$. Do the numbers of the diagonal follow any order? Would it be $\begin{pmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 1\end{pmatrix}$ or $\begin{pmatrix}2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ Thank you!

4. The order would depend on how you make your matrix P. The eigenvalues will be in the same column as the corrisponding eigenvector. both are similar to the original.

5. My eigenvector for 2 is <-1, 1, 0> if I'm right and the two eigenvectors for 1 are <-2, 1, 0> and <1, 0, 1>. So I think my diagonal matrices may be wrong even though I'm leaning towards the first one with the 2 in the center.

6. Originally Posted by jennifer1004
My eigenvector for 2 is <-1, 1, 0> if I'm right and the two eigenvectors for 1 are <-2, 1, 0> and <1, 0, 1>. So I think my diagonal matrices may be wrong even though I'm leaning towards the first one with the 2 in the center.
Your eigenvalues and vectors are correct .

What I am saying is this

If you make your matrix P such that $PAP^{-1}=D$

like this this

$\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

$\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

$\begin{bmatrix}
1 & -1 & -2 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
\end{bmatrix}
$

Then D would look like this

$\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.