Find Eigenvalues and Eigenvectors of: $$A=\begin{pmatrix} 4 & 0 & -9 \\ 9 & -5 & -8\\ 0 & 0 &-5 \end{pmatrix}$$
$$A=\begin{vmatrix} 4-\lambda & 0 & -9 \\ 9 & -5-\lambda & -8\\ 0 & 0 &-5-\lambda \end{vmatrix}=(-5-\lambda)\begin{vmatrix} 4-\lambda & 0 \\ 9 & -5-\lambda \\ \end{vmatrix}=(-5-\lambda)^2(4-\lambda)$$
So $\lambda_{1}=-5$ and $\lambda_{2}=4$
$$\begin{pmatrix} 4+5 & 0 & -9 \\ 9 & -5+5 & -8\\ 0 & 0 &-5+5 \end{pmatrix}=\begin{pmatrix} 9 & 0 & -9 \\ 9 & 0 & -8\\ 0 & 0 &0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1\\ 0 & 0 &0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1\\ 0 & 0 &0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 &1 \end{pmatrix}$$
So the eigenvectors of $\lambda_{1}=-5$ are : $$\{\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\}$$
$$\begin{pmatrix} 4-4 & 0 & -9 \\ 9 & -5-4 & -8\\ 0 & 0 &-5-4 \end{pmatrix}=\begin{pmatrix} 0 & 0 & -9 \\ 9 & -9 & -8\\ 0 & 0 &-9 \end{pmatrix}=\begin{pmatrix} 0 & 0 & 1 \\ 1 & -1 & 0\\ 0 & 0 &0 \end{pmatrix}=\begin{pmatrix} 1 & -1 &0 \\ 0 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}$$
So the eigenvectors of $\lambda_{2}=4$ are : $$\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\,\begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}\}$$
Where did I get it wrong? How can it be that I got a matrix with a rank of $2$ but just $1$ solution?
Even more odd here are some solutions from calculators that give diffreant results:
$\endgroup$ 71 Answer
$\begingroup$The mistake is here ($\lambda=4$): you have to solve $$I) x_1 + -x_2 + 0 = 0$$ $$II) 0 + 0 + 0 = 0$$ $$III) 0 + 0 + x_3 = 0$$ therefore $x_1=x_2, x_3=0$ and $$ker \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} = $$ $$\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix}\}$$
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