How to find nonsingular matrix?

**jennifer1004** · April 20th 2009, 07:04 AM

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} $

**TheEmptySet** · April 20th 2009, 07:24 AM

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.

**jennifer1004** · April 21st 2009, 02:24 PM

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.

**TheEmptySet** · April 21st 2009, 02:28 PM

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

**jennifer1004** · April 21st 2009, 04:46 PM

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}$ ?

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