Linear Transformations and Reflections

Question

I have the following exercise: Let $T:\mathbb{R}^2\ \rightarrow\mathbb{R}^2$ be a linear transformation that first performs a horizontal shear that transforms the standard unit vectors $e_1, e_2$. First, we turn $e_2$ into $e_2 + 2e_1$ leaving $e_1$ unchanged, and then we reflect it through the line $x_1=- x_2$.

Find the standard matrix of $T$.

So I'm kind of confused about this one, I would think that the transformation itself (before reflecting it), would be $$\begin{pmatrix} 1&2\\ 0&1\\ \end{pmatrix}$$

The reflection is unclear to me. If $x_1 = -x_2$, do we interchange them or just turn $x_1$ into $-x_2$ like so? $$\begin{pmatrix} 0&-1\\ 0&1\\ \end{pmatrix}$$

If instead we do interchange them, shouldn't it then be the following?:

$$\begin{pmatrix} 0&-1\\ 1&2\\ \end{pmatrix}$$

Instead the answer is

$$\begin{pmatrix} 0&-1\\ -1&-2\\ \end{pmatrix}$$

Which I really don't understand. Would appreciate some help, thanks in advance!

Answer 1

As the columns of the matrix of a linear map are the images of the vectors of the basis, the matrix of the shear is

$$S=\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}.$$

Now let's have a look at the reflection $R$. The vector $a= (1, -1)^T$ belongs to the reflection line $L$. It is therefore unchanged under $R$. While the vector $b=(1,1)^T$ is orthogonal to $L$ and is therefore mapped into $R(b)=-b=(-1,-1)^T$.

However, we need to find the images under $R$ of $e_1$ and $e_2$. We have

$$e_1 = \frac{1}{2}\left(a+b\right), \, e_2 = \frac{1}{2}\left(b-a\right)$$ and

$$R(e_1) = \frac{1}{2}\left(R(a)+R(b)\right) = -e_2, \, R(e_2) = \frac{1}{2}\left(R(b)-R(a)\right) = -e_1.$$ Based on that, the matrix of $R$ is

$$R=\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}.$$

And the one of $T$

$$T=RS=\begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & -1\\ -1 & -2 \end{pmatrix}.$$

Note: you made an error in the matrix of the reflection. A reflection is invertible. Hence no non-vector can have the zero vector for image.

Answer 2

The standard matrix of the plane linear transformation $T$ satisfying $T(e_{1}) = (a, c)$ and $T(e_{2}) = (b, d)$ has the image vectors as columns: $$ \left[\begin{array}{@{}cc@{}} a & b \\ c & d \\ \end{array}\right]. $$ The natural way to proceed is to find (probably by geometry) the images of the standard basis under the given reflection. (The same idea gives the shear matrix, but you've already found that successfully.)