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

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

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.

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 $\displaystyle e_n$ I just don't know how...

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

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.

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.

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;

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

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, $\displaystyle T(x_1 , x_2 , x_3)=...$ 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 $\displaystyle T(e_1)=b_3$? And I would do that by multiplying $\displaystyle M$ and $\displaystyle e_1$? Or,

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

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

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.

[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 $\displaystyle T(e_1)=b_3$?

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.

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.

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.