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

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 $\displaystyle P^{-1}AP=\mbox{diag}(0,-2)$

Re: Finding diagonalization of a matrix

Thank you FernandoRevilla for checking my work. I must have been over thinking it!