Finding $\theta$ and unit vector for a reflection matrix

I'm given a transformation matrix, and am told that it is a reflection combined with scaling:

\begin{pmatrix} 4 & 3 \\ 3 & -4 \end{pmatrix}

I'm asked to do two things:

1) Find $\theta$: how do I understand an angle theta in relation to a reflection? This is intuitive to me for a rotation, but not for a reflection. Is it twice the angle between a given vector and the line of reflection, so that the reflection across the line is viewed as a rotation? How might I find this?

2) Find the unit vector for the line of reflection. Since this is a reflection, I set $2u_1^2 - 1 = 4$ to get $u_1 = \pm\frac 52$, but then $2u_2^2 - 1 = -4$ nets an imaginary number. What am I doing wrong? I have heard that reflections involve imaginary numbers and we haven't covered that yet, but I'm lost on how else I might find the unit vector.

I'm looking for pointers/concept explanations, rather than outright answers. Appreciate any help!

• Where did the expression $2u_1^2 - 1 = 4$ come from? – amd Sep 21 '17 at 21:31

The line of reflection bisects the angle between a vector and its image. Pick any convenient unit vector $\mathbf u$ and compute its reflection $\mathbf u'$. The vector $\mathbf u+\mathbf u'$ bisects the angle between them (remember the parallelogram rule for vector addition) and so lies along the line of reflection. I hope that you can find the angle this vector makes with the $x$-axis and derive a unit vector from it on your own.
Points on the line of reflection are mapped to themselves by this transformation, so in the language of eigenvectors, it is the eigenspace of $1$.
The eigenvalues are $-5,5$, and this tells you that the matrix consists of a scaling by $5$, and of a reflection, since (putting apart the scaling) there is a vector that remains the same and another that is inverted.
The eigenvector corresponding to $\lambda=5$ is $(3,1)$, and so that is the axis of reflection.