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Assume that T is a linear transformation. Find the standard matrix of T. $T:\mathbb{R}^2 \rightarrow\mathbb{R}^2$ first reflects points through the line $x_2$=$x_1$ and then reflects points through the horizontal $x_1$-axis.

My Solution , that is incorrect :- The standard matrix for the reflection through the line $x_2$=$x_1$ is \begin{bmatrix}0&1\\1&0\end{bmatrix}

The standard matrix for the reflection through the horizontal $x_1$-axis is : \begin{bmatrix}1&0\\0&-1\end{bmatrix}

When we multiply this we get : \begin{bmatrix}0&-1\\1&0\end{bmatrix}

This answer is being rejected. Can you please advise me what am I doing wrong? Thank you.

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  • $\begingroup$ Sir, I thank you for your patience. I am not clear about multiplying this by $x_1$,$x_2$. I did multiply it row x column times, and found the answer \begin{bmatrix}0&1\\-1&0\end{bmatrix} $\endgroup$ – SubbaKudmi May 26 '17 at 2:20
  • $\begingroup$ Sir, that is \begin{bmatrix}x_2\\-x_1\end{bmatrix} Sorry I realize my mistake, I have corrected it. But I am still not clear, as this was not the original question. Thank you for your help. $\endgroup$ – SubbaKudmi May 26 '17 at 2:54
  • $\begingroup$ Restoring my original comment: It appears you multiplied in the wrong order. The transformations should be multiplied so that the left to right order is the reverse of the time order (i.e., the first transformation to be performed goes on the right, and the last one goes on the left). $\endgroup$ – quasi May 26 '17 at 3:59
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The task is simple . . .

Let $B$ be the matrix for the transformation which reflects a point about the line $x_2=x_1$. Then $$B = \begin{bmatrix}0&1\\1&0\end{bmatrix}$$

Let $A$ be the matrix for the transformation which reflects a point about the $x_1$-axis. Then $$A = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ The matrix for the transformation which first reflects a point about the line $x_2=x_1$, and then reflects the result about the $x_1$-axis is just $$AB = \begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$

To test it, let $v={\displaystyle{\begin{bmatrix}x_1\\x_2\end{bmatrix}}}$.

First apply the transformations by hand . . .

  • Start with $$\begin{bmatrix}x_1\\x_2\end{bmatrix}$$
  • Next, reflect about the line $x1=x2$: $$\begin{bmatrix}x_1\\x_2\end{bmatrix} \rightarrow \begin{bmatrix}x_2\\x_1\end{bmatrix}$$
  • Next, reflect the above result about the $x_1$-axis: $$\begin{bmatrix}x_2\\x_1\end{bmatrix} \rightarrow \begin{bmatrix}x_2\\-x_1\end{bmatrix}$$

Next, check to see if $(AB)v$ yields the same result: $$(AB)v = \begin{bmatrix}0&1\\-1&0\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}x_2\\-x_1\end{bmatrix}$$ so it checks.

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  • $\begingroup$ Sir, Thank you very much. That is indeed the answer. Can we clean up some of the chat, so that it helps others. Also is the question purposefully reversed so that it will yield the same answer, no matter which transformation is performed first ? I thank you for your help. $\endgroup$ – SubbaKudmi May 26 '17 at 3:34
  • $\begingroup$ Yes, you can delete all your comments, and I'll delete mine. $\endgroup$ – quasi May 26 '17 at 3:36
  • $\begingroup$ @SubbaKudmi: No, the order counts. If you do the transformations in the reverse order (.i.e., $BA$ instead of $AB$), you won't get the same matrix, and you won't get the required result if you apply it to the vector $\displaystyle{\begin{bmatrix}x_1\\x_2\end{bmatrix}}$. $\endgroup$ – quasi May 26 '17 at 3:54

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