My professor didn't go over this in class at all. I was looking online and found some solutions, but was not able to get a $2\times 2$ matrix in any of the examples.
The formula I was using was $v (V^t)/\|v\|^2$. I do not understand how the transpose helps though. I kept getting the answer:
First row first column: $6/\sqrt{61}$ First row, second column: Second row, first column: $5/\sqrt{61}$ Second row, second column:
I am looking for more clarification or maybe a different formula.
On
Plug the vector that defines the line into your formula, remembering that $v$ is a column vector there: $$\frac1{61}\pmatrix{6\\5} \pmatrix{6&5}.$$ The product of the $2\times1$ matrix $v$ and $1\times2$ matrix $v^T$ is a $2\times2$ matrix, just as you’ve been trying to find. (Note, by the way, that you need the square of the norm of $v$, not its norm, in this formula.)
With that out of the way, note that the result is not the reflection that you’re looking for, but instead the orthogonal projection of a vector onto that line. You can, however, use that projection matrix to construct the reflection matrix: the reflection can be performed by reversing the component of the vector being reflected that’s orthogonal to the line. The above projection gives you a way to decompose any vector into a component parallel to the line—the projection of the vector—and one orthogonal to the line.
Let $R$ be that reflection. Then $R(6,5)=(6,5)$ and, since $(-5,6)$ is orthogonal to $(6,5)$, $R(-5,6)=-(-5,6)=(5,-6)$. So, the matrix of $R$ with respect to the basis $\bigl((6,5),(-5,6)\bigr)$ is $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$. Now, let $P=\left[\begin{smallmatrix}6&-5\\5&-6\end{smallmatrix}\right]$; that is, the columns of $P$ are the vectors of the basis. Then the matrix that you're after is$$P.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.P^{-1}=\frac1{61}\begin{bmatrix}11&60\\60&-11\end{bmatrix}.$$