Finding a linear transformation.

[peripatein]

Hi,

Homework Statement

How may I find (or prove that there isn't) a linear transformation which satisfies T: R³->R₁[x], ker T = Sp{(1,0,1), (2,-1,1)}?

Homework Equations

The Attempt at a Solution

I am not sure how to approach this. I understand that kerT is the group of all vectors (x,y,z) in R³ so that T(x,y,z) = 0 = Sp{(1,0,1), (2,-1,1)}. So x = alpha, y = -beta, z = alpha + beta? Hence, x,y,z=0? Hence, T has to be one-to-one? Since dim(R₁[x]) is 2, does that mean that there is no such linear transformation?

 

[mfb]

I don't understand what you try to do with your alpha and beta.
Every vector which can be written in that way is part of ker T - and now? What about vectors which cannot be written like that?

Since dim(R₁[x]) is 2, does that mean that there is no such linear transformation?

You can use a dimensional analysis to prove that there is no surjective linear transformation (with the given ker T), but that is not relevant here.

 

[peripatein]

Won't dimensional analysis show that dimIm(T) must be equal to 1? How does that help me disprove that no such linear transformation exists?

 

[mfb]

Why do you think you can disprove it at all?

dim Im(T)=1, I agree

 

[peripatein]

I don't know, wrong inference from your previous post I presume. In any case, how shall I proceed?

 

[HallsofIvy]

R_1[x] is the set of all polynomials of the form ax+ b for real numbers a and b (unless I am misunderstanding- I would call that R[x] without the "1"). We can represent that by the column vector \begin{bmatrix}a \\ b \end{bmatrix} and so we can represent a linear transformation from R^3 to R_1[x] by a 3 by 2 matrix:
\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}p \\ q\end{bmatrix}

Saying that the kernel is spanned by (1, 0, 1) and (2, -1, 1) means that
\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}a+ c \\ d+ e\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}
and
\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}2 \\ -1\\ 1\end{bmatrix}= \begin{bmatrix}2a- b+ c \\ 2d- e+ f\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}

So we must have a+ c= 0, d+ e= 0, 2a- b+ c= 0, and 2d- e+ f= 0. That gives four equations with which we can solve for four of a, b, c, d, e, and f in terms of the other two.

 

[peripatein]

Would you care to explain the logic behind such approach? For instance, why did you choose a matrix whose first two elements in the first row are the elements of the column vector [a,b]?

 

[mfb]

a to f are just some constants (with arbitrary names), as this 2x3-matrix represents an arbitrary linear transformation from R^3 to R^2 (or R[x] in this special case).

 

[peripatein]

Right, but it is for a reason that a and b appear in both cases, namely first two elements in the first row of the matrix AND the elements of the column vector [a,b]. I am asking why is that.
Moreover, I have a general question - are all different bases linearly dependent?

 

[mfb]

No, they are not related. As you can see, HallsofIvy replaced the names by p and q in the first equation to avoid collisions.

 

[peripatein]

Okay. The result I obtained was a=b=-c and d=-e=-f/3.
How do I derive my answer from these values?

 

[mfb]

As you have no additional restrictions, you can choose two values for those parts to find a linear transformation.
a=b=-c=5, d=e=-f/3=7 for example.

 

[peripatein]

Hence, how should the final answer be presented? Shouldn't it be presented in the form of a polynomial ax+b?

 

[mfb]

The final answer should give some unique way to map all vectors (g,h,i) to vectors px+q.

 

[peripatein]

In that case, I don't think I understand. I thought I was to provide a general term for a transformation answering the criteria as presented in the question. Would you clarify, please?

 

[mfb]

You have to give a single linear transformation - this is a specific way to map every vector of R^3 to a polynomial: Like an equation T(g,h,i)=... which can be used for all values of g,h,i and gives a result which depends on g,h,i.

 

[peripatein]

Let me see if I got it correctly. Supposing I take the values for a,b,c,d,e and f, and substitute them into the original matrix as suggested by HallsofIvy, I get a single column vector whose elements are 5x+5y-5z and 7x+7y-21z. To what end does that serve me?