# How to find nonsingular matrix?

• Apr 20th 2009, 07:04 AM
jennifer1004
How to find nonsingular matrix?
Find, if possible, a nonsingular matrix P such that $P^{-1}AP$ is diagonal.

$
\begin{pmatrix}1 & 1 & 2\\0 & 1 & 0\\0 & 1 & 3\end{pmatrix}
$
• Apr 20th 2009, 07:24 AM
TheEmptySet
Quote:

Originally Posted by jennifer1004
Find, if possible, a nonsingular matrix P such that $P^{-1}AP$ is diagonal.

$
\begin{pmatrix}1 & 1 & 2\\0 & 1 & 0\\0 & 1 & 3\end{pmatrix}
$

First you need to find the eigenvalues so you can find the eigen Basis for the matrix (if it exists)

$
\begin{pmatrix}1-\lambda & 1 & 2\\0 & 1-\lambda & 0\\0 & 1 & 3 -\lambda \end{pmatrix}
$

Expanding along the first column we get

$(1-\lambda)[(1-\lambda)(3-\lambda)-(1)(0)]=(1-\lambda)^2(3-\lambda)$

So the two eigenvalues are $\lambda =1,\lambda=3$

Using $\lambda =1$ we get the matrix

$
\begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 1 & 2 \end{pmatrix}
$

Reducing we get

$
\begin{pmatrix}0 & 1 & 2\\0 & 0 & 0\\0 & 0 & 0 \end{pmatrix}
$

This implies that there are two parameters in the solution so let

$x=s, z=t$ so the last equation is

$y+2z=0 \iff y+2t=0 \implies y =-2t$

so the solution is

$\begin{pmatrix}
s \\
-2t \\
t
\end{pmatrix} =
s \cdot \begin {pmatrix}
1 \\ 0 \\ 0
\end {pmatrix}
+t \cdot \begin {pmatrix}
0 \\ -2 \\ 1
\end{pmatrix}
$

So these are the first two eigenvectors

Now using the eigen values $\lambda =3$ we get

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

Reducing we get

$
\begin{pmatrix}1 & 0 & -1\\0 & 1 & 0\\0 & 0 & 0 \end{pmatrix}
$

This has 1 parameter in its solution so let $z=t$

We can see that $y=0$ and $x=z \iff x=t$

So the solution is

$\begin{pmatrix}
t \\ 0 \\ t
\end{pmatrix}=t \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end {pmatrix}$

The eigenvectors are the columns of the matrix P.
• Apr 21st 2009, 02:24 PM
jennifer1004
so my answer is simply $
\begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}
$
?

Thank you for showing your work. It has helped me so much. I can now use your example to help me figure out problems on my own.
• Apr 21st 2009, 02:28 PM
TheEmptySet
Quote:

Originally Posted by jennifer1004
so my answer is simply $
\begin{pmatrix}1 & 0 & 1\\0 & -2 & 0\\0 & 1 & 1\end{pmatrix}
$
?

Thank you for showing your work. It has helped me so much. I can now use your example to help me figure out problems on my own.

Yes that is the solution
• Apr 21st 2009, 04:46 PM
jennifer1004
Out of curiosity, you wouldn't multiply the eigenvector for $\lambda=3$ by 3 for the final matrix, which would make it $
\begin{pmatrix}1 & 0 & 3\\0 & -2 & 0\\0 & 1 & 3\end{pmatrix}$
?