### Instruction: Find the matrix that corresponds to projecting onto the vector $\begin{pmatrix} 1 \\ 7 \end{pmatrix}.$.### Response: To project onto a vector, I need to find a matrix that maps every vector onto its component in the direction of that vector.
This means that the matrix should satisfy the equation $P \begin{pmatrix} x \\ y \end{pmatrix} = \dfrac{\begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 7 \end{pmatrix}}{\begin{pmatrix} 1 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 7 \end{pmatrix}} \begin{pmatrix} 1 \\ 7 \end{pmatrix}.$
This is the formula for the scalar projection times the direction vector.
To find the matrix P, I can plug in some basis vectors and see what they get mapped to.
For example, if I plug in $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, I get $P \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \dfrac{1}{50} \begin{pmatrix} 1 \\ 7 \end{pmatrix}$.
This means that the first column of P is $\dfrac{1}{50} \begin{pmatrix} 1 \\ 7 \end{pmatrix}$.
Similarly, if I plug in $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, I get $P \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \dfrac{7}{50} \begin{pmatrix} 1 \\ 7 \end{pmatrix}$.
This means that the second column of P is $\dfrac{7}{50} \begin{pmatrix} 1 \\ 7 \end{pmatrix}$.
Therefore, the matrix P is $\dfrac{1}{50} \begin{pmatrix} 1 & 7 \\ 7 & 49 \end{pmatrix}$.