# Finding diagonalization of a matrix

• Sep 25th 2012, 06:15 AM
nivek0078
Finding diagonalization of a matrix
Thanks to all that can help. I'm given the matrix A= [-3, square root of 3] [ negative square root of 3, 1]. The question asks you to find the diagonalized matirx.

I find the determinant to be A(A+2) where lamba=A so A1=0, A2=-2. Is this right? If it is what is my P matrix? I have only one eigenvector from A2=-2 to be [square root 3][1] and with that you can't get a P inverse. Which is needed for the formula D=P^1AP
• Sep 25th 2012, 07:37 AM
FernandoRevilla
Re: Finding diagonalization of a matrix
Quote:

Originally Posted by nivek0078
I'm given the matrix A= [-3, square root of 3] [ negative square root of 3, 1]. I find the determinant to be A(A+2) where lamba=A so A1=0, A2=-2. Is this right?

Right.

Quote:

If it is what is my P matrix? I have only one eigenvector from A2=-2 to be [square root 3][1] and with that you can't get a P inverse. Which is needed for the formula D=P^1AP
Verify that a basis of $\displaystyle V_0$ is $\displaystyle \{(1,\sqrt{3})\}$ and a basis of $\displaystyle V_{-2}$ , $\displaystyle \{(\sqrt{3},1)\}$ so, $\displaystyle P=\begin{pmatrix} 1&\sqrt{3}\\\sqrt{3}&1\end{pmatrix}$ satisfies $\displaystyle P^{-1}AP=\mbox{diag}(0,-2)$
• Sep 25th 2012, 09:10 AM
nivek0078
Re: Finding diagonalization of a matrix
Thank you FernandoRevilla for checking my work. I must have been over thinking it!