Find the matrix A of the reflection in the line $\mathbb{R}^2$ that consists of all scalar multiples of the vector $(6,5)$. 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.

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$$P.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.P^{-1}=\frac1{61}\begin{bmatrix}11&60\\60&-11\end{bmatrix}.$$
