Problem finding the transformation in this case

[subopolois]

Homework Statement

give the matrix in the transformation T in each case:
T: is a reflection in the line y= -x

Homework Equations

none that i know of for this question

The Attempt at a Solution

i know y= -x goes from the top left to bottom right in a cartesian plane, so the reflection of the line with initial point (1,1) in the first quadrant would end up in the third quadrant. i get my answer as:
-1 0
0-1
but the answer in the back of the book is:
0 -1
-1 0
why is their matrix the exact opposite as mine?

 

[gabbagabbahey]

Well, what does the vector(1,1) become once it is reflected along the line y=-x?

 

[subopolois]

Well, what does the vector(1,1) become once it is reflected along the line y=-x?

it becomes (-1,-1)

 

[gabbagabbahey]

Good, that means that:

$$T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

How would you go about solving this for T?

 

[subopolois]

Good, that means that:

$$T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

How would you go about solving this for T?

im not sure with this one,wait would you just rearrange to find t?

 

[subopolois]

Good, that means that:

$$T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

How would you go about solving this for T?

but how does it end up being:
0 -1
-1 0
and not
-1 0
0 -1
?

 

[gabbagabbahey]

There is more than one method, but I would write:

$$T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} T_{11} & T_{12} \\ T_{21} & T_{22} \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

Then carry out the matrix multiplication and solve for the components of T...if you want to use this method, you will also need another equation of the same form...what does T do to the vector (-1,1)?

 

[subopolois]

There is more than one method, but I would write:

$$T \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} T_{11} & T_{12} \\ T_{21} & T_{22} \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 \\ -1 \end{pmatrix}$$

Then carry out the matrix multiplication and solve for the components of T...if you want to use this method, you will also need another equation of the same form...what does T do to the vector (-1,1)?

i still don't see how since the point (1,1) is transformed to (-1,-1) it gives
0 -1
-1 0
when i see this it says to me, that y= -1 from the first column when i think it should be x=-1 in the first column and likewise with the second column

 

[gabbagabbahey]

Well, what is T suuposed to do to the vector (-1,1)...does your T actually do that?

Since there are 4 unique elements to T, you need four equations to solve for them...looking at the effect of T on a single vector only gives you two equations, so you need to consider the effect on two vectors.

 

[HallsofIvy]

If A is a linear transformation from vector space V to itself and {e₁, e<sub>2<sup>, ..., en} is a basis for V, you can find the matrix representation for A by applying to each of the basis vectors in turn. For example, suppose the matrix representation of A, in that basis, is
$$\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right]$$
then, since the array representation of the basis vectors is just [1, 0, 0], [0, 1, 0], [0, 0 , 1], Applying A to e1 is just
$$\left[\begin{array}{ccc}a_{11} & a_{12} & a{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a{32} & a_{33}\end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\left]= \left[\begin{array}{c} a_{11} \\ a_{21} \\ a_{31}\end{array}\right]$$
That is, applying A to e1 tells us what the first column is, e2, the second column, etc.

The standard basis for R2 is {[1 0], [0 1]}. Applying A to [1 0] gives [0 -1] and applying A to [0 1] is [0 -1] and to [0 1], [-1 0] so the matrix is
$$\left[\begin{array}{cc} 0 & -1 \\-1 & 0\end{array}\right]$$

However, if we use basis {[0 1],[1 0]}, that is the same vectors but in different order, that swaps the columns for A:
$$\left[\begin{array}{cc} -1 & 0 \\0 & -1\end{array}\right]$$

In the first example, e1= [1 0], e2= [0 1]. In the second example, e1= [0 1], e2= [1 0].

Same vector space, same linear transformation, same vectors in each basis but, because the order or basis vectors is different, different bases and so different matrices.</sup></sub>