<s>[INST] Properties of eigenvectors $$A :=\begin{pmatrix}  
          5 &−6 &0\\
         2 &−2& 0  \\
          0 &0 &2  \\
\end{pmatrix} $$
Determine an eigenvector of the matrix that is perpendicular to the vector $\mathbf v = (1, −4, 1).$
I think I need to first determine the eigenvalues and eigenvectors of $A$. Then I can use that a linear combination of eigenvectors of eigenvalue $λ$ is again an eigenvector of eigenvalue $λ$. How can I solve it?
 [/INST] From the definition of the eigenvector $\mathbf v$ corresponding to the eigenvalue $λ$ we have
$$A\mathbf v=λ\mathbf v$$
Then:
$$A\mathbf v-λ\mathbf v=(A-λI)\mathbf v=0$$
Equation has a nonzero solution if and only if
$$\det(A-λI)=0$$
$$\det(A-λI)=\begin{vmatrix} 5-λ& -6& 0 \\  2& -2-λ& 0\\ 0& 0& 2-λ\end{vmatrix} $$$$=-λ^3+5λ^2-8λ+4=-(λ-1)(λ^2-4λ+4)=-(λ-1)(λ-2)^2=0$$
$\lambda_1=1$ and $\lambda_2=2$ with multiplicity $2$ are the eigenvalues.
For each eigenvalue, look for the associated eigenvector. $( A' − I ) \mathbf X = \mathbf{0}$. If $\lambda_1=1$ then $$\begin{pmatrix}
 4& -6& 0 \\
  2& -1& 0\\
 0 & 0 & 1
\end{pmatrix}\cdot \begin{pmatrix}
 x \\
  y\\
 z
\end{pmatrix}=\begin{pmatrix}
 0 \\
  0\\
 0
\end{pmatrix}$$
Solving the system you will find a vector of components, for example, $\mathbf v_1=(x_1,y_1,z_1)\ne \mathbf 0$. If $\mathbf v_1\cdot \mathbf v =0$ then $\mathbf v_1 \perp \mathbf v$. Same steps for $\lambda_2=2$.
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