I'm curious why a linear operator T on R$^2$ is a reflection if it has orthogonal eigenvectors v$_1$ and v$_2$ with eigenvalues 1 and -1.
I'm hoping someone could show me a linear transformation that has eigenvectors with eigenvalues +1 and -1 but please choose a reflection other than just over the x-axis or y-axis. A reflection about some line would be great (the line should be one eigenvector from my understanding). I propose the orthogonal vectors [1,2] and [2,-1]
Thanks!
On
Theo's answer settles the general question. I'll look at the particular one. We want $$\pmatrix{a&b\cr c&d\cr}\pmatrix{1&2\cr2&-1}=\pmatrix{1&-2\cr2&1}$$ as that matrix takes $(1,2)$ to $(1,2)$, and $(2,-1)$ to $-(2,-1)=(-2,1)$. We find an inverse: $$\pmatrix{1&2\cr2&-1}^{-1}={1\over5}\pmatrix{1&2\cr2&-1}$$ so $$\pmatrix{a&b\cr c&d\cr}={1\over5}\pmatrix{1&-2\cr2&1}\pmatrix{1&2\cr2&-1}={1\over5}\pmatrix{-3&4\cr4&3\cr}$$
On
The quick description is $$ \left( \begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{array} \right) $$
As in Gerrry's answer, when the elements of the matrix are rational, the angle $\theta$ will not be familiar, it will be $\arctan$ of something, possibly negating or adding an integer multiple of $\pi$
Judging by your comment, your own concept of a reflection seems to be an image similar to reflecting in the $x$ or $y$ axis, but somehow the picture is rotated so that the axis in question becomes some arbitrary line. This is something we can work with.
We'll need the idea of an orthogonal operator. An operator $U$ is orthogonal if $$(Ux) \cdot (Uy) = x \cdot y$$ for all $x, y \in \Bbb{R}^2$. In particular, note that $$\|Ux\|^2 = (Ux) \cdot (Ux) = x \cdot x = \|x\|^2,$$ hence $U$ preserves distances. We can also recover the angle between vectors using the dot product, hence $U$ must also preserve angles. This means that the orthogonal operators are essentially just rotations and reflections. So, if you're thinking about rotating a picture so that one axis becomes an arbitrary line, orthogonal operators are what you need.
It also follows that orthogonal operators must be injective, since $$U(x) = 0 \implies \|U(x)\| = 0 \implies \|x\| = 0 \implies x = 0.$$ Since $U$ is an injective operator on a finite-dimensional space, this implies $U$ is invertible. Inverses of orthogonal operators are also orthogonal, as $$(U^{-1} x) \cdot (U^{-1} y) = (U(U^{-1} x)) \cdot (U(U^{-1}y)) = x \cdot y.$$
Now, let's normalise the eigenvectors $v_1$ and $v_2$. Let $$u_i = \frac{v_i}{\|v_i\|}.$$ Then $u_1$ is a unit eigenvector corresponding to eigenvalue $1$, and similarly for $u_2$ with $-1$. Note that $u_1 \perp u_2$, and hence they form an orthonormal basis.
Consider the unique linear map $U$ taking $(1, 0)$ to $u_1$, and $(0, 1)$ to $u_2$. I claim that $U$ is an orthogonal operator. Given any $(a, b), (c, d) \in \Bbb{R}^2$, \begin{align*} U(a, b) \cdot U(c, d) &= U(a(1, 0) + b(0, 1)) \cdot U(c(1, 0) + d(0, 1)) \\ &= (aU(1, 0) + bU(0, 1)) \cdot (cU(1, 0) + dU(0, 1)) \\ &= (au_1 + bu_2) \cdot (cu_1 + du_2) \\ &= ac(u_1 \cdot u_1) + ad(u_1 \cdot u_2) + bc(u_2 \cdot u_1) + bd(u_2 \cdot u_2) \\ &= ac + 0 + 0 + bd \\ &= (a, b)\cdot(c,d). \end{align*}
Let $R$ be the reflection in the $x$-axis, i.e. $R(x, y) = (x, -y)$. Let us follow, intuitively speaking, the composition $$U R U^{-1}.$$ First, what does $U^{-1}$ do? Remember, it is an orthogonal operator, and will be some kind of reflection/rotation, preserving distances and angles. Note that $U$ maps the $x$-axis (i.e. $\operatorname{span}(1, 0)$) to the line spanned by $u_1$, i.e. the eigenspace of $T$ corresponding to $1$. Also, $U^{-1}$ maps the $y$-axis to the eigenspace of $T$ corresponding to $-1$. Hence, $U^{-1}$ will do the opposite: it will map $\operatorname{span}(u_1) \mapsto \operatorname{span}(1, 0)$ and $\operatorname{span}(u_2) \mapsto \operatorname{span}(0, 1)$.
So, when we apply $U R U^{-1}$, we apply $U^{-1}$ first, rotating the whole picture so that the eigenspace corresponding to $1$ maps to the $x$-axis, and the eigenspace corresponding to $-1$ maps to the $y$-axis. Then, $R$ is applied, leaving the $x$-axis fixed, but flipping the sign of the $y$-coordinate; i.e. the actual reflection has taken place. Finally, we apply $U$, undoing the effects of $U^{-1}$. The now reflected vector is rotated back, nullifying the change in perspective effected by $U^{-1}$.
In total, $U R U^{-1}$ has reflected the vector in the line $\operatorname{span}(v_1)$.
Now, it's not hard to see that this operator is indeed $T$. In fact, if you simply show that these linear operators agree on a basis, then they agree everywhere. The most sensible basis is $(u_1, u_2)$, the orthonormal basis of eigenvectors. By definition of eigenvectors, we have \begin{align*} T(u_1) = 1u_1 = u_1 \\ T(u_2) = (-1)u_2 = -u_2. \end{align*} On the other hand, we have \begin{align*} URU^{-1}(u_1) = UR(1, 0) = U(1, 0) = u_1 \\ URU^{-1}(u_2) = UR(0, 1) = U(0, -1) = -u_2. \end{align*} Therefore, the two operators agree on a basis, so they are the same operator, i.e. $$T = URU^{-1}.$$ Hopefully my arguments are sufficient to convince you that $URU^{-1}$ is a reflection in the way that you understand the word, and hence so is $T$.