1. **Problem statement:** Given the vector $v = \begin{bmatrix}-1 \\ 1 \\ 2\end{bmatrix}$, compute the Householder matrix $H_v$ and determine the effect of multiplying $H_v$ by $v$. 2. **Formula for Householder matrix:** The Householder matrix $H_v$ is defined as $$ H_v = I - 2 \frac{vv^T}{v^Tv} $$ where $I$ is the identity matrix, $v$ is the given vector, $v^T$ is its transpose, and $v^Tv$ is the dot product of $v$ with itself. 3. **Calculate $v^Tv$:** $$ v^Tv = (-1)^2 + 1^2 + 2^2 = 1 + 1 + 4 = 6 $$ 4. **Calculate the outer product $vv^T$:** $$ vv^T = \begin{bmatrix}-1 \\ 1 \\ 2\end{bmatrix} \begin{bmatrix}-1 & 1 & 2\end{bmatrix} = \begin{bmatrix}1 & -1 & -2 \\ -1 & 1 & 2 \\ -2 & 2 & 4\end{bmatrix} $$ 5. **Compute $H_v$:** $$ H_v = I - 2 \frac{vv^T}{6} = I - \frac{1}{3} vv^T $$ So, $$ H_v = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} - \frac{1}{3} \begin{bmatrix}1 & -1 & -2 \\ -1 & 1 & 2 \\ -2 & 2 & 4\end{bmatrix} = \begin{bmatrix}1 - \frac{1}{3} & 0 + \frac{1}{3} & 0 + \frac{2}{3} \\ 0 + \frac{1}{3} & 1 - \frac{1}{3} & 0 - \frac{2}{3} \\ 0 + \frac{2}{3} & 0 - \frac{2}{3} & 1 - \frac{4}{3}\end{bmatrix} $$ Simplifying, $$ H_v = \begin{bmatrix}\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & -\frac{1}{3}\end{bmatrix} $$ 6. **Effect of multiplying $H_v$ by $v$:** Calculate $$ H_v v = \left(I - 2 \frac{vv^T}{v^Tv}\right) v = v - 2 \frac{v (v^T v)}{v^T v} = v - 2v = -v $$ This means multiplying $v$ by $H_v$ reflects $v$ to its negative. **Final answer:** The Householder matrix is $$ H_v = \begin{bmatrix}\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} & -\frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & -\frac{1}{3}\end{bmatrix} $$ and multiplying $H_v$ by $v$ results in $$ H_v v = -v $$ which is the reflection of $v$ about the hyperplane orthogonal to $v$.