Simple reflection of vector in $\mathbb{R}^2$, weird answer

Question

"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

Answer 1

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$$\begin{bmatrix}3&4\\4&-3\end{bmatrix}.\begin{bmatrix}1&0\\0&-1\end{bmatrix}.\begin{bmatrix}3&4\\4&-3\end{bmatrix}^{-1}=\begin{bmatrix}-\frac7{25}&\frac{24}{25}\\\frac{24}{25}&\frac7{25}\end{bmatrix}.$$

Answer 2

Notice that if $A$ is a reflection, then $A^2=I$. It is not an error to find $A(25,0)$ if you're trying to find what maps to $(25,0)$, because that is the answer! $A(A(25,0))=A^2(25,0)=(25,0)$.

Answer 3

The author’s “obvious” error isn’t: every reflection is its own inverse, so $\mathcal Z^{-1}(25,0)=\mathcal Z(25,0).$ I would hope that this property of reflections was mentioned somewhere earlier in the text so that it wasn’t simply pulled out of thin air for this particular exercise.

Now, you can construct the matrix of a linear transformation from the images of an arbitrary basis, but you end up with a matrix that’s expressed relative to that basis or, as you did, a matrix that has different bases for its “input” and “output.” For the matrix you constructed, we have $\mathcal Z_M(1,0)=(3,4)$ and $\mathcal Z_m(0,1)=(4,-3)$: the outputs coordinates are expressed relative to the standard basis, but the inputs are relative to your convenient-for-this-reflection basis. We want the standard basis on both sides, so you need to apply a change of basis on the right. The correct matrix of the reflection is therefore $$\begin{bmatrix}3&4\\4&-3\end{bmatrix}\begin{bmatrix}3&-4\\4&3\end{bmatrix}^{-1} = \frac1{25}\begin{bmatrix}-7&24\\24&7\end{bmatrix}.$$