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:
http://bit.ly/2c1J7oq
Best Answer
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 $$ker \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} = $$ $$\{\begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix}\}$$