# Thread: Linear Transform Question that has me a bit confused.

1. ## Linear Transform Question that has me a bit confused.

I'm sure this is a super easy question but I'm not entirely sure what these people are talking about in this section.

Let $\varepsilon =\{e_1 , e_2, e_3\}$ be a standard basis for $\mathbb{R}{^3}$. Let $\beta=\{b_1 , b_2, b_3\}$ for a vector space V. And let $T:\mathbb{R}\rightarrow{V}$ be a linear transform with the property that $T(x_1, x_2, x_3)=(2x_3 -x_2)b_1 -(2x_2)b_2+(x_1+3x_3)b_3$
a. Compute $T(e_1), T(e_2), T(e_3)$.
b. Compute $[T(e_1]_\beta,[T(e_2)]_\beta$, and $[T(e_3)]_\beta$.
c. Find the matrix for $T$ relative to $\varepsilon$ and $\beta$

OK I understand all of the assumptions with the exception of $T(x_1, x_2, x_3)=(2x_3 -x_2)b_1 -(2x_2)b_2+(x_1+3x_3)b_3$ which I can't really make sense of at the moment.

2. ## Re: Linear Transform Question that has me a bit confused.

Hey bkbowser.

Basically the transformation takes the vector X and turns it into B. Since it is a linear there exists a matrix A such that f(X) = AX = B. To get you started on the first question consider that e1 = <1,0,0>, e2 = <0,1,0> and e3 = <0,0,1>.

I would suggest that you find the matrix that takes X and creates B via the matrix A. Just match up the components for every element of B by putting them in the right positions for the matrix A and it should all make sense.

3. ## Re: Linear Transform Question that has me a bit confused.

Originally Posted by chiro
Hey bkbowser.

Basically the transformation takes the vector X and turns it into B. Since it is a linear there exists a matrix A such that f(X) = AX = B. To get you started on the first question consider that e1 = <1,0,0>, e2 = <0,1,0> and e3 = <0,0,1>.

I would suggest that you find the matrix that takes X and creates B via the matrix A. Just match up the components for every element of B by putting them in the right positions for the matrix A and it should all make sense.
For a, I already know that ever I have to do to get the correct solution is going to involve the standard vector $e_n$ I just don't know how...

Like at the moment I have no clue how to get from, $T(x_1, x_2, x_3)=(2x_3 -x_2)b_1 -(2x_2)b_2+(x_1+3x_3)b_3$ to the conclusion that $T(e_1)=b_3 ... T(e_2)=-b_1-2b_2 ... T(e_3)= 2b_1 +3b_3$

The only thing I can think of algebraically is to distribute the 3 vectors into the scalar expressions and that doesn't seem to get me anything that makes any more sense then what I have already.

4. ## Re: Linear Transform Question that has me a bit confused.

So basically your first component will have 2x3 - x2 which means the first row of your matrix A will be [0 -1 2] and when you multiply that by [x1 x2 x3]' (i.e. transpose) you will get for your first component 0*x1 - 1*x2 + 2*x3 = 2x3 - x2. You can do the same thing for the other two rows and you have your A matrix which represents your linear transformation.

Now given this you can just plug in values for the vector X and you can even look at inverse transformations and change of basis.

5. ## Re: Linear Transform Question that has me a bit confused.

Originally Posted by chiro
So basically your first component will have 2x3 - x2 which means the first row of your matrix A will be [0 -1 2] and when you multiply that by [x1 x2 x3]' (i.e. transpose) you will get for your first component 0*x1 - 1*x2 + 2*x3 = 2x3 - x2. You can do the same thing for the other two rows and you have your A matrix which represents your linear transformation.

Now given this you can just plug in values for the vector X and you can even look at inverse transformations and change of basis.
OK that makes sense.

So my matrix looks like;

$\begin{smallmatrix} 0&0&1\\ -1&-2&0\\0&0&3 \end{smallmatrix}$

This matrix looks terrible...

Sorry I basically have no information from the text on this from this section on this specific problem as it doesn't solve a concrete example of it. And the notation is different enough that I can't relate it to an example problem. I'll try and attach something that might make explaining this to me easier.

So, $T(x_1 , x_2 , x_3)=...$ relates to equation (1)?

And the matrix I just wrote with your help is M? Or equation (4) if you prefer.

Now I need to figure out that $T(e_1)=b_3$? And I would do that by multiplying $M$ and $e_1$? Or,

$[\begin{smallmatrix} 0&0&1\\ -1&-2&0\\0&0&3 \end{smallmatrix}] [\begin{smallmatrix} 1\\0\\0 \end{smallmatrix}]=$

And of course this looks wrong... since it nets me a zero vector...

6. ## Re: Linear Transform Question that has me a bit confused.

That is not quite right: the matrix should be:

[0 -1 2]
[0 -2 0]
[1 0 3]

You should expand out the matrix when multiplied against the column vector [x1 x2 x3]' for your own verification.

7. ## Re: Linear Transform Question that has me a bit confused.

[0 -1 2]
[0 -2 0]
[1 0 3]

.....[0].........[-1] ........[2]
x_1[0] + x_2[-2] + x_3[0] = 1x_1 + - 1x_2 - 2x_2 + 2x_3 + 3x_3
....[1] ........[0] ..........[3]

OK this makes a lot of sense to me since it just looks like a linear combination of vectors. But how am I supposed to know how to go from the Right Hand Side to the Left Hand Side like that? I think the row order of the matrix you gave me doesn't matter. So any possible combination of rows is just as good as another? And I can't row reduce it or change it's form?

And even if that's right I still don't see how to get $T(e_1)=b_3$?

8. ## Re: Linear Transform Question that has me a bit confused.

The order matters because it has to correspond to the right basis. Your b1,b2,b3 has to match up with your x1,x2,x3. If you change a row in the matrix then you are changing the basis vectors that are being mapped to.

In terms of T(e1) we apply Ax where x = <1,0,0> which gives <0*1 -1*0 + 2*0, 1*0 -2*0 + 0*0, 1*1 + 0*0 + 3*0>' = <0,0,1> = b3. The same sort of calculation holds for other arbitrary vectors.

9. ## Re: Linear Transform Question that has me a bit confused.

Originally Posted by chiro
The order matters because it has to correspond to the right basis. Your b1,b2,b3 has to match up with your x1,x2,x3. If you change a row in the matrix then you are changing the basis vectors that are being mapped to.

In terms of T(e1) we apply Ax where x = <1,0,0> which gives <0*1 -1*0 + 2*0, 1*0 -2*0 + 0*0, 1*1 + 0*0 + 3*0>' = <0,0,1> = b3. The same sort of calculation holds for other arbitrary vectors.
OK so A is

[0 -1 2]
[0 -2 0]
[1 0 3]

And that's the matrix of transformation for T?

And T(e1) is just the product of the matrix A and the standard basis vector e1.

T(e1) we apply Ax where x = <1,0,0> which gives <0*1 -1*0 + 2*0, 1*0 -2*0 + 0*0, 1*1 + 0*0 + 3*0>' = <0,0,1> = b3

The part I don't understand is how that equals b3? To me b3 doesn't look like it equals anything but itself.

10. ## Re: Linear Transform Question that has me a bit confused.

The b vectors b1,b2,b3 are just basis vectors. We are mapping R3 to R3 here which means that our basis vectors are <1,0,0>, <0,1,0>, and <0,0,1> for e1,e2,e3 and b1,b2,b3. We could choose them to be other basis vectors but these are the assumed ones if not stated otherwise.

b3 corresponds to a vector <0,0,1> and a general vector in the space of the b's is d*b1 + e*b2 + f*b3 just as the original vector is g*e1 + h*e2 + i*e3 for scalars d,e,f,g,h,i.