Linear transformations question

[reflex]

Hi all,

So this question is fairly basic, but I want to be certain I have the right idea before I do the other parts (asks about it in standard basis etc). It's a book question:

Homework Statement

Here are the vectors : u=[ 1 2 0] v=[2 5 0] w=[1 1 1]
This forms a basis B of R3 B={u,v,w}

D is a transformation from R3 to R3
D(u)=2v
D(v)=3w
D(w)=0

Homework Equations

So it asks me to put D in matrix form in the B coordinate system:

The Attempt at a Solution

4 3 0
10 3 0
0 3 0

Is this it? If not a hint of how to think about it would be greatly appreciated.
Sorry about the lack of good formatting.

 

[jbunniii]

Here are the vectors : u=[ 1 2 0] v=[2 5 0] w=[1 1 1]
This forms a basis B of R3 B={u,v,w}

D is a transformation from R3 to R3
D(u)=2v
D(v)=3w
D(w)=0

Let's start with D(u) = 2v. What are the coordinates of u and 2v in the B coordinate system?

 

[lanedance]

I don't think so, how did you get that matrix? Have u tried applying the matrix to the basis vectors to see that it's satisfies the definition?

 

[reflex]

@jbunii - Since v is in basis B, I just doubled v to get 2v. Does that make sense?

@LaneDance - When you multiply D times [1 0 0] (only as a vector) you get the first column of D, which I thought was 2v. So it does if I'm correct about 2v..

 

[reflex]

@LaneDance - I got the matrix by having D= [ D(u) D(v) D(w) ] . Basically I'm asking if that is the correct method. Thanks!

 

[HallsofIvy]

Yes, that is the correct method but that does NOT give the matrix you show.

In general, to represent a linear transformation A, from vector space U to vector space V, as a matrix, using ordered basis B_U for U, ordered basis B_V for V, apply A to each of the basis vectors in B_u in turn, writing the result in terms of B_V. The coefficients are the numbers in each column of the matrix representation.

Here, D is a linear transformation for R³ to itself and we are using the same basis for both domain and range. D(u)= 2v= 0u+ 2v+ 0w, D(v)= 3w= 0u+ 0v+ 3w, and D(w)= 0= 0u+ 0v+ 0w. Those will be the three columns of the matrix representation of D.

 

[reflex]

Aha! Thank you very much, that makes much more sense than what I was doing. The output should be what mine was in standard, not B!

Thank you very much.

 

[Ray Vickson]

Aha! Thank you very much, that makes much more sense than what I was doing. The output should be what mine was in standard, not B!

Thank you very much.

You need to be careful about exactly which matrix you are talking about: is it the matrix of the basis transformation, or is it the basis of the component transformation? Here is what I mean, illustrated in a little example. Suppose e1 and e2 are the standard unit vectors along the two axes in a 2-dimensional vector space. If T(e1) = e1 + e2, and T(e2) = 2e1, we can write
T(e_1) = [1\:,\:1]\begin{pmatrix}e_1\\e_2\end{pmatrix},\;<br /> T(e_2) = [2\:, \: 0] \begin{pmatrix}e_1\\e_2\end{pmatrix} \Longrightarrow <br /> M = \begin{pmatrix} 1&amp;1\\2&amp;0\end{pmatrix}, where M is the matrix of the transformation of the basis vectors:
T\pmatrix{e_1\\e_2} = <br /> \begin{pmatrix} 1&amp;1\\2&amp;0\end{pmatrix} \pmatrix{e_1\\e_2}.

On the other hand, the vector
x = x_1 e_1 + x_2 e_2 transforms as
T(x) = x_1 T(e_1) + x_2 T(e_2) = (x_1 + 2x_2) e_1 + x_1 e_2<br /> \Longrightarrow T\pmatrix{x_1\\x_2} = \begin{pmatrix}1&amp;2\\1&amp;0\end{pmatrix}\pmatrix{x_1\\x_2} , which gives the matrix
M&#039; = \begin{pmatrix}1&amp;2\\1&amp;0\end{pmatrix} = M^T. So, would you be talking about M or about M'?

RGV

 

[reflex]

I don't know exactly what you are referring to when you say component, but here is what I meant:

The output from my original(incorrect) matrix was the B vectors in the standard basis.
So to get the output of [4 10 0]' (2v in standard), I would need to multiply B*D*inverse(b) by the vector u ([1 2 0]').

Caveat: I'm a linear newbie.

Thank you everyone for all the input so far!

 

[Ray Vickson]

I don't know exactly what you are referring to when you say component, but here is what I meant:

The output from my original(incorrect) matrix was the B vectors in the standard basis.
So to get the output of [4 10 0]' (2v in standard), I would need to multiply B*D*inverse(b) by the vector u ([1 2 0]').

Caveat: I'm a linear newbie.

Thank you everyone for all the input so far!

Components are like the x_1 and x_2 in my example. They are the coefficients of e_1 and e_2 when you express the 2-dimensional vector x in terms of the two basis vectors e_1 and e_2. In about 99% of the uses of matrices in linear algebra, we want the matrix of transformations of the components (i.e., the matrix M'). Very occasionally, we want the matrix of the basis transformations (the matrix M). So again: which one do you want?

RGV

 

[reflex]

Components are like the x_1 and x_2 in my example. They are the coefficients of e_1 and e_2 when you express the 2-dimensional vector x in terms of the two basis vectors e_1 and e_2. In about 99% of the uses of matrices in linear algebra, we want the matrix of transformations of the components (i.e., the matrix M'). Very occasionally, we want the matrix of the basis transformations (the matrix M). So again: which one do you want?

RGV

I want the coefficients to be in basis B, not E. So I guess the solution provided above does that, since inputting [1 0 0]' (using B basis, this is u) gives me back [0 2 0] (which is 2v in B basis).

The question was: "Find the matrix D in the B coordinate system". Unfortunately it does not get more specific. The above answer seems to do that however.