Standard matrix for reflection across the line y=-x

[fattycakez]

Homework Statement

Let T : R²→R², be the matrix operator for reflection across the line L : y = -x

a. Find the standard matrix [T] by finding T(e1) and T(e2)

b. Find a non-zero vector x such that T(x) = x

c. Find a vector in the domain of T for which T(x,y) = (-3,5)

Homework Equations

The Attempt at a Solution

a. I found [T] =
0 -1
-1 0

b. I'm not really sure what this is asking. Do I just pick a random x = (x₁,x₂)
and then plug in T(x) = x?

c. T(-3, 5) would be (-5, 3). Is that what the question is asking? Is (-5, 3) in the domain of T?

Help me, I'm stupid :D

 

[fourier jr]

a. I got $\Big( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \Big)$

b. you want to find a vector that the matrix doesn't move. So you don't pick a random one, you start with T(x, y) = (x, y) & then find a vector that satisfies that equation.

c. that's right

 

[LCKurtz]

Homework Statement

Let T : R²→R², be the matrix operator for reflection across the line L : y = -x

a. Find the standard matrix [T] by finding T(e1) and T(e2)

b. Find a non-zero vector x such that T(x) = x

c. Find a vector in the domain of T for which T(x,y) = (-3,5)

Homework Equations

The Attempt at a Solution

a. I found [T] =
0 -1
-1 0

I agree.

b. I'm not really sure what this is asking. Do I just pick a random x = (x₁,x₂)
and then plug in T(x) = x?

If you think geometrically about the reflection you are given, isn't it pretty clear which points wouldn't move?

c. T(-3, 5) would be (-5, 3). Is that what the question is asking? Is (-5, 3) in the domain of T?

Yes.

 

[fattycakez]

Thanks guys!

a. I got $\Big( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \Big)$

b. you want to find a vector that the matrix doesn't move. So you don't pick a random one, you start with T(x, y) = (x, y) & then find a vector that satisfies that equation.

c. that's right

How did you get that matrix?

I agree.
If you think geometrically about the reflection you are given, isn't it pretty clear which points wouldn't move?
Yes.

After fiddling with some numbers to try to get it to work I got T(-1, 1) = (-1, 1) which would be a point on the line y = -x so I guess all the points on the line wouldn't move obviously right?

 

[LCKurtz]

After fiddling with some numbers to try to get it to work I got T(-1, 1) = (-1, 1) which would be a point on the line y = -x so I guess all the points on the line wouldn't move obviously right?

Fiddling?? Do you mean guessing? Experimenting?? What happens to any point on the line $y=-x$ when you reflect it in that line? Or if you solve$$
\begin{bmatrix}
0& -1\\
-1 & 0
\end{bmatrix}
\begin{bmatrix} a\\b\end{bmatrix} =
\begin{bmatrix} a\\b\end{bmatrix}
$$for $a$ and $b$? No guesswork needed.

 

[fattycakez]

Fiddling?? Do you mean guessing? Experimenting?? What happens to any point on the line $y=-x$ when you reflect in in that line? Or if you solve$$
\begin{bmatrix}
0& -1\\
-1 & 0
\end{bmatrix}
\begin{bmatrix} a\\b\end{bmatrix} =
\begin{bmatrix} a\\b\end{bmatrix}
$$for $a$ and $b$? No guesswork needed.

It reflects onto the line but in the opposite quadrant?

 

[LCKurtz]

No. Think of your line as a mirror. Any point some distance away on one side of the mirror reflects to appear the same distance away on the other side of the mirror. What about a paint dot right on the mirror. Where does its reflection appear to be?

 

[fattycakez]

No. Think of your line as a mirror. Any point some distance away on one side of the mirror reflects to appear the same distance away on the other side of the mirror. What about a paint dot right on the mirror. Where does its reflection appear to be?

It doesn't move then right? Its reflection is on its original location?

 

[fourier jr]

How did you get that matrix?

Never mind I did the matrix for the wrong transformation it made sense at the time anyway

 

[HallsofIvy]

Here's how I would do that problem: Any 2 by 2 matrix can be written as $$\begin{bmatrix}a & b \\ c & d \end{bmatrix}$$. "Reflecting about the line y= -x" the vector <1, 0> is mapped into < 0, -1> and the vector <0, 1> is mapped into <-1, 0>.

So we must have $$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix} 0 \\ -1\end{bmatrix}$$ and $$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\ 0\end{bmatrix}$$.

Doing the matrix multiplications on the left and setting the components equal to the right gives you four equations to solve for a, b, c, and d.

 

[LCKurtz]

It doesn't move then right? Its reflection is on its original location?

Right. So for that reflection, it's pretty obvious geometrically which points in $R^2$ give $T(x) = x$.

 

[fattycakez]

Right. So for that reflection, it's pretty obvious geometrically which points in $R^2$ give $T(x) = x$.

Cool thanks, that makes sense! I learned everything I know about math from U of A, its not my fault :D