Please correct my work, finding Eigenvector

Question

Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.

$$A= \begin{bmatrix} +3 & +2\\ -2 & -3\\ \end{bmatrix} $$

$$A-\ell I = \begin{bmatrix} +3-\ell & +2\\ -2 & -3-\ell\\ \end{bmatrix} $$ Then Determinant should be zero : $$ \begin{vmatrix} +3-\ell & +2\\ -2 & -3-\ell\\ \end{vmatrix}=(3-\ell)(-3-\ell)+4=0 \to \ell^2-5=0 \to \ell_{1}=\sqrt 5,\ell_{2}=-\sqrt5 , $$ $$ \begin{bmatrix} 3-\ell_{1}& +2\\ -2 & -3-\ell{1}\\ \end{bmatrix}= \begin{bmatrix} 3-\sqrt 5=0.76& +2\\ -2 & -3-\sqrt 5=-5.24\\ \end{bmatrix}\to{R_1\Leftarrow\Rightarrow R2 \mapsto} \begin{bmatrix} -2 & -5.24\\ 0.76& +2\\ \end{bmatrix}\to{R_1=R1/{-2}\mapsto} \begin{bmatrix} 1 & 2.62\\ 0.76& +2\\ \end{bmatrix}\to{R_2=R2-0.76R1\mapsto} \begin{bmatrix} 1 & 2.62\\ 0 & 0.009\\ \end{bmatrix}\to{R_2=R2/0.009\mapsto} \begin{bmatrix} 1 & 2.62\\ 0 & 1\\ \end{bmatrix}\to{R_1=R1-2.62R1\mapsto} \begin{bmatrix} 1 & 0 | 0\\ 0 & 1 | 0\\ \end{bmatrix} $$ $then$ $$ V_1= \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$ I am stuck here , matrix of $0,0$ is the right answer Eigenvector 1 ?

Answer 1

The eigenvalues are correct. However you go wrong in the computation of the eigenvectors:

$$ \begin{bmatrix} 3-\sqrt{5} & 2\\ -2 & -3-\sqrt{5} \end{bmatrix} $$

Divide the first row by $3-\sqrt{5}$, which is the same as multiplying it by $(3+\sqrt{5})/2$, getting

$$ \begin{bmatrix} 1 & \frac{3+\sqrt{5}}{2}\\ -2 & -3-\sqrt{5} \end{bmatrix} $$

Now adding to the second row the first one multiplied by $2$ brings the matrix in the form

$$ \begin{bmatrix} 1 & \frac{3+\sqrt{5}}{2}\\ 0 & 0 \end{bmatrix} $$

so you know that one eigenvector is

$$ \begin{bmatrix} -\frac{3+\sqrt{5}}{2}\\ 1 \end{bmatrix} $$

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The computations for the other eigenvalue are similar

$$ \begin{bmatrix} 3+\sqrt{5} & 2\\ -2 & -3+\sqrt{5} \end{bmatrix} $$ $$ \begin{bmatrix} 1 & \frac{3-\sqrt{5}}{2}\\ -2 & -3+\sqrt{5} \end{bmatrix} $$

$$ \begin{bmatrix} 1 & \frac{3-\sqrt{5}}{2}\\ 0 & 0 \end{bmatrix} $$

So the other eigenvector you're looking for is $$ \begin{bmatrix} -\frac{3-\sqrt{5}}{2}\\ 1 \end{bmatrix} $$

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Of course a matrix that diagonalizes $A$ is

$$ P= \begin{bmatrix} -\frac{3+\sqrt{5}}{2} & -\frac{3-\sqrt{5}}{2}\\ 1&1 \end{bmatrix} $$

Answer 2

The problem is that you rounded. The 0.009 in the second row is in fact a 0.