# How to find the standard matrix of a shear / reflection transformation from $\mathbb R^2$ to $\mathbb R^2$

Let $$T : \Bbb{R}^2 \rightarrow \Bbb{R}^2$$ be the transformation that first performs a horizontal shear so that $$e_2 \rightarrow e_2 + 2e_1$$ (leaving $$e_1$$ unchanged) and then reflects the points through the line $$y = x$$

(a) Find the standard matrix $$A$$ for $$T$$

(b) Find the standard matrix for the inverse mapping directly by finding a transformation that first undoes the reflection through the line y=x and then undoes the horizontal shear. Show that this matrix is the same as the matrix you found for $$A^{-1}$$

I'm on (a) and have

$$\underbrace{ \begin{bmatrix} x_1 & y_1\\ x_2 & y_2\\ \end{bmatrix} }_{M}$$ $$\xrightarrow{(2)e_1 + e_2 \rightarrow e_2} \begin{bmatrix} x_1 & y_1\\ 2x_1+x_2 & 2y_1+y_2\\ \end{bmatrix}$$

$$\begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix} \begin{bmatrix} x_1 & y_1\\ 2x_1+x_2 & 2y_1+y_2\\ \end{bmatrix} = \begin{bmatrix} 2x_1+x_2 & 2y_1+y_2\\ x_1 & y_1\\ \end{bmatrix} = T$$

(a) $$A = T \cdot M^{-1}$$

$$A=\begin{pmatrix}2x_{1+x_2}&2y_1+y_2\\ \:\:\:\:x_1&y_1\end{pmatrix}\begin{pmatrix}\frac{y_2}{x_1y_2-y_1x_2}&-\frac{y_1}{x_1y_2-y_1x_2}\\ \:\:\:\:-\frac{x_2}{x_1y_2-y_1x_2}&\frac{x_1}{x_1y_2-y_1x_2}\end{pmatrix}$$ $$= \begin{pmatrix}\frac{2x_{1+x_2}y_2-x_2\left(2y_1+y_2\right)}{y_2x_1-y_1x_2}&\frac{-2x_{1+x_2}y_1+x_1\left(2y_1+y_2\right)}{y_2x_1-y_1x_2}\\ 1&0\end{pmatrix}$$

However the result for $$A$$ seems wonky. Am I on the right track?

• To obtain the standard matrix for a linear transformation, put the images of the basis vectors in the columns Commented May 26, 2021 at 20:58
• I'm sorry I'm not understanding what you mean. I have the matrix M is that the basis vector in this case? the images being the T? which columns? I think I'm not setting something up correctly and the A = T * M^-1 is not the correct approach. Commented May 26, 2021 at 21:13
• What is $T(e_1)$ and $T(e_2)$? Commented May 26, 2021 at 21:43

The standard matrix for $$T$$ is $$A=RS$$, where $$S=\pmatrix{1&2\\0&1}$$ and $$R=\pmatrix{0&1\\1&0}$$.

Can you take it from here?

• yes, I was able to find the correct method. I understand your comment now. I was just not contextualizing it correctly. Thank you! Commented May 27, 2021 at 0:04
• Hey J.W. Tanner, do R and S stand for Reflection and Shear in your response here? Commented Jun 6 at 18:17
• @DavidD.: I suppose so Commented Jun 6 at 19:36

I was not on the right track at all. I needed to set up using an Identity matrix. the solution is as follows.

$$T(\vec{x})=A\vec{x}=A_2A_1\vec{x}$$

$$I= \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}$$

$$T_1 (e_1)=(e_1)$$

$$T_1 (e_2)=(e_2)+2(e_1)$$

$$A_1= \begin{bmatrix} 1 & 2\\ 0 & 1\\ \end{bmatrix}$$

$$T_2 (e_1)=(e_2)$$

$$T_2 (e_2)=(e_1)$$

$$A_2= \begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix}$$

$$T(\vec{x})=T_2(T_1(\vec{x}))=A_2A_1\vec{x}$$

$$A= \begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 0 & 1\\ 1 & 2\\ \end{bmatrix}$$