# Diagonal Matrix

• April 20th 2009, 07:20 AM
jennifer1004
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}
$
• April 20th 2009, 07:32 AM
TheEmptySet
Quote:

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.
• April 21st 2009, 12:35 PM
jennifer1004
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!
• April 21st 2009, 12:44 PM
TheEmptySet
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.
• April 21st 2009, 12:59 PM
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.
• April 21st 2009, 01:24 PM
TheEmptySet
Quote:

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 (Clapping).

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.