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}

$

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- Apr 20th 2009, 07:20 AMjennifer1004Diagonal Matrix
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}

$ - Apr 20th 2009, 07:32 AMTheEmptySet
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. - Apr 21st 2009, 12:35 PMjennifer1004
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!

- Apr 21st 2009, 12:44 PMTheEmptySet
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.

- Apr 21st 2009, 12:59 PMjennifer1004
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.

- Apr 21st 2009, 01:24 PMTheEmptySet
Your eigenvalues and vectors are correct (Clapping).

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.