### Instruction: A projection takes $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$ to $\begin{pmatrix} \frac{60}{13} \\ \frac{12}{13} \end{pmatrix}.$  Which vector does the projection take $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ to?### Response: This is a problem about projections of vectors onto some subspace.
To find the vector that the projection takes $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ to, I need to know what subspace the projection is projecting onto.
One way to find the subspace is to use the fact that the projection of a vector is parallel to the subspace.
This means that the vector $\begin{pmatrix} \frac{60}{13} \\ \frac{12}{13} \end{pmatrix}$ is parallel to the subspace that the projection is projecting onto.
To find a vector that is parallel to this one, I can multiply it by any scalar.
For convenience, I will choose the scalar 13, so that the vector becomes $\begin{pmatrix} 60 \\ 12 \end{pmatrix}$.
This vector is in the subspace that the projection is projecting onto, so I can use it as a basis vector for the subspace.
Now, I need to find the projection of $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ onto the subspace spanned by $\begin{pmatrix} 60 \\ 12 \end{pmatrix}$.
One way to do this is to use the formula for the projection of a vector u onto a vector v, which is $$\text{proj}_v u = \frac{u \cdot v}{v \cdot v} v$$ where $\cdot$ denotes the dot product of two vectors.
In this case, u is $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ and v is $\begin{pmatrix} 60 \\ 12 \end{pmatrix}$.
I can calculate the dot products as follows: $$u \cdot v = (-2)(60) + (2)(12) = -120 + 24 = -96$$ $$v \cdot v = (60)(60) + (12)(12) = 3600 + 144 = 3744$$
Then, I can plug these values into the formula and simplify: $$\text{proj}_v u = \frac{u \cdot v}{v \cdot v} v = \frac{-96}{3744} \begin{pmatrix} 60 \\ 12 \end{pmatrix} = -\frac{1}{39} \begin{pmatrix} 60 \\ 12 \end{pmatrix} = \begin{pmatrix} -\frac{60}{39} \\ -\frac{12}{39} \end{pmatrix} = \begin{pmatrix} -\frac{20}{13} \\ -\frac{4}{13} \end{pmatrix}$$