Question
Show that $\mathbf{x}=\left(\begin{array}{l}1 \\ 1\end{array}\right)$ is a null direction for $K=\left(\begin{array}{rr}1 & -2 \\ -2 & 3\end{array}\right)$, but $\mathbf{x} \notin \operatorname{ker} K$.
Step 1
A vector $\mathbf{x}$ is a null direction for $K$ if $K \mathbf{x}$ is a scalar multiple of $\mathbf{x}$. Mathematically, this means there exists a scalar $\lambda$ such that $K \mathbf{x} = \lambda \mathbf{x}$. Show more…
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