"The linear mapping $\mathcal Z$ denotes reflecting a vector about an axis (line) with direction vector $\mathbf a=(3,4)$ in the plane. Write down the components of $\mathbf y= \mathcal Z\mathbf x$ for an arbitrary $\mathbf x=(x_1,x_2)$ and find a particular vector $\mathbf y$, which gets mapped to $(25,0)$."
Straighforward approach, isn't it:
Using $(3,4)$ and a vector orthogonal to it as a basis (edit 1: not using a standard basis aka the unit vectors) for $\mathbb{R}^2$, we can write the matrix representation of $\mathcal Z$ as $$
\mathcal Z_M = \begin{bmatrix}
a_1 & a_2 \\
a_2 & -a_1 \\
\end{bmatrix} = \begin{bmatrix}
3 & 4 \\
4 & -3 \\
\end{bmatrix}
$$
The matrix-vector product of $\mathcal Z_M$ with arbitrary vector $\mathbf x$ yields
$$ \mathbf y = \begin{bmatrix}
3x_1 + 4x_2 \\
4x_1 – 3x_2 \\
\end{bmatrix}
$$
Finding what maps to $(25,0)$ is a matter of solving
$$ \left[
\begin{array}{cc|c}
3&4&25\\
4&-3&0
\end{array}
\right] $$
which yields $\mathbf y = (3,4)$.
Except it's the wrong result, or it's equivalent to the correct one for reasons I can't see, or it's a technical error for which I've lost an hour of my time:
The author says, for the first part, $\mathbf y = {\frac{1}{25}}(-7x_1+24x_2, 24x_1 + 7x_2)$ and for the second $\mathbf x = \mathcal Z(25,0)=(-7,24)$. Okay, so for the second part he's obviously made an error and mapped $(25,0)$ via the matrix instead of looking what maps to it. However, I can't see how that first part is correct by any stretch of the imagination, except, perhaps by using a different set for the basis.
Is there a basis which maybe is implicit when finding matrix representations of these simple geometric transformations in linear algebra?
Edit 2:
So, I basically thought I can construct corresponding matrices for mappings by taking ANY base, not just ${{(1,0), (0,1)}}$ by adjusting the following theorem, which only seems to work for $\mathbf {e_1, e_2} … \mathbf{e_n}$
src: http://linear.ups.edu/html/section-LT.html
Best Answer
You know that $\mathcal Z(3,4)=(3,4)$ and that $\mathcal Z(4,-3)=(-4,3)$. That is, the matrix of $\mathcal Z$ with respect to the basis $\bigl\{(3,4),(4,-3)\}$ is $\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$. So, the matrix of $\mathcal Z$ with respect to the standard basis is$$\begin{bmatrix}3&4\\4&-3\end{bmatrix}.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.\begin{bmatrix}3&4\\4&-3\end{bmatrix}^{-1}=\begin{bmatrix}-\frac7{25}&\frac{24}{25}\\\frac{24}{25}&\frac7{25}\end{bmatrix}.$$