0
$\begingroup$

Let $M:\mathbb{R}^2\ \rightarrow\mathbb{R}^2$ be the linear transformation that first reflects it through the line $x_1=x_{2}$, and then rotates each point counterclockwise around the origin by $\frac{5 \pi}{4}$ radians.

Find the standard matrix of $M$

So I was thinking, since it's the line $x_1 = x_2$, looking at it graphically, I would have

$$\begin{pmatrix} cos(x)&-sin(x)\\ sin(x)&cos(x)\\ \end{pmatrix}$$ which then becomes

$$\begin{pmatrix} sin(x)&cos(x)\\ cos(x)&-sin(x)\\ \end{pmatrix}$$

which then becomes (using $x = \frac{5\pi}{4}$) $$\begin{pmatrix} -\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\ \end{pmatrix}$$

But that's the wrong answer. Would really appreciate some help with these, I don't fully understand how to algebraically find / convert these reflections yet.

Thanks!

$\endgroup$

1 Answer 1

1
$\begingroup$

The reflection through the line $x_1=x_2$ maps $\left(1,1\right)$ into itself and it maps $\left(-1,1\right)$ into $\left(1,-1\right)$. So, its matrix with respect to the standard basis is $\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$. And the matrix of a counterclockwise rotation around the origin with angle $\frac{5\pi}4$ is$$\begin{bmatrix}-\frac1{\sqrt2}&\frac1{\sqrt2}\\-\frac1{\sqrt2}&-\frac1{\sqrt2}\end{bmatrix}.$$So, take$$\begin{bmatrix}-\frac1{\sqrt2}&\frac1{\sqrt2}\\-\frac1{\sqrt2}&-\frac1{\sqrt2}\end{bmatrix}.\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}\frac1{\sqrt2}&-\frac1{\sqrt2}\\-\frac1{\sqrt2}&-\frac1{\sqrt2}\end{bmatrix}.$$

$\endgroup$
10
  • $\begingroup$ Why is the first row, second column the one that is positive rather than second row second column? We are first supposed to reflect, correct? If so, $-sin(x)$ will be in the second row second column I believe? $\endgroup$
    – JakeDrone
    Jun 6, 2021 at 23:23
  • $\begingroup$ Because it's the matrix$$\begin{bmatrix}\cos\left(\frac{5\pi}4\right)&-\sin\left(\frac{5\pi}4\right)\\\sin\left(\frac{5\pi}4\right)&\cos\left(\frac{5\pi}4\right)\end{bmatrix}.$$ $\endgroup$ Jun 7, 2021 at 6:15
  • 1
    $\begingroup$ No, I am not reflecting first. The matrix that I wrote is the matrix of a counterclockwise rotation around the origin with angle $\frac{5\pi}4$. After having computed that matrix, I multiply it by the matrix corresponding to the reflection. $\endgroup$ Jun 7, 2021 at 13:35
  • 1
    $\begingroup$ In this problem, it is stated that the first transformation is the reflection. In that other problem, the reflection is the second transformation. $\endgroup$ Jun 7, 2021 at 21:57
  • 1
    $\begingroup$ At no point I wrote that I first take the counterclockwise rotation and then the reflection. Take a look at my answer. I computed the matrix $R$ corresponding to the reflection, I computed the matrix $O$ corresponding to the counterclockwise rotation and then I wrote that the answer is $O.R$. By this order. The matrix $R$ is on the right since it is the first operation and the matrix $R$ is on the left since it is the second operation. The operations are written from the right to the left. $\endgroup$ Jun 14, 2021 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.