# Homework Help: Question regarding linear transformations

1. Mar 14, 2006

### Mathman23

Hi the following assigment.

Given $$P_{2} (D)$$ be a vector space polynomials of at most degree n=2.

Looking at the transformation $$T: P_2(D) \rightarrow D^2$$, where

T(p) = [p(-i),p(i)].

1) Show that this transformation is linear.

I order to show this I hold my transformation up against the definition of the generel linear transformation?

2) The base $$B = (1,x,x^2)$$ in $$P_{2} (D)$$, and the standard base $$B' = (e_1,e_2)$$ in $$D^2$$. Find the matrix representation for T in relation to the bases.

Since T(p) = [p(-i), p(i)], then there most be two polynomials of degree 2 which represent T:

$$T(p(-i),p(i)) = $\left( \begin{array}{c} 1 + i + i^2 & \\ 1 +(-i) + (-i)^2 & \\ \end{array}$$ right? Then the standard matrix representation for the transformation must be the matrix: $$A(p(i)) = \[ \left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 & \\ 0 & 0 & 0 & 1 & 0 & \\ 0 & 1 & 0 & 0 & 0 & \\ 0 & 1 & 0 & 0 & 0 & \\ 1 & 1 & 3 & 0 & 0 & \\ \end{array} \right]$$$

/Fred

Last edited: Mar 14, 2006
2. Mar 14, 2006

### AKG

To check that T is linear, find out what the definition of "linear transformation" is and then verify that T is such a thing. For number 2, the whole thing doesn't make sense. T takes polynomials as arguments, whereas you're writing T(p(i), p(-i)), i.e. you're giving T an argument from D². Now T is a transformation from a 3 dimensional space to a two dimensional space, so you should get a 2x3 matrix. What is this 5x5 matrix, where is it coming from? And again, why are you putting p(i) in the argument of A? Do you understand at all what's going on?

3. Mar 15, 2006

### Mathman23

(Urgend)

Dear AKG,

I think I get (2) now.

The linear mapping is given as follows

$$T: P_2(C) \rightarrow C^2$$ which defined by $$T(p) = [p(-i), p(i)]$$

Isn't that equivalent to:

$$T(p(a_0 + a_1 + {a_2} ^2)) = [p((a_0 + a_1(-i) + {a_2}(-i)^2), p((a_0 + a_1(i) + {a_2}(i)^2)]$$

There the transformation takes a polynomial of at most degree 2 and gives the solutions for original polymial?

If my assumption regarding the polynomials are correct, then what about the two bases

B = (1,x, x^2) for P_2(C) and B'= (e1,e2) for C^2 ??

Here I'm suppose to find the matrixrepresentation in relation to the bases.

If I insert the base B into the polynomial on the leftside I the following three polymials

T(p(1)) = 3

T(p(x)) = x^2 + 2x

T(p(x^2)) = x^4 + 2x^2

What do I do with these polynomials?

Sincerley
Fred

Last edited: Mar 15, 2006
4. Mar 15, 2006

### AKG

What is p(a0 + a1 + a2²)? It doesn't really make sense.
What do you mean "solutions"?
What is p(x²)? Do you know what a polynomial is? And why are you getting polynomials as your values when you apply T to, well, whatever it is you're applying it to? T maps into C², so you should get an ordered pair (a,b) where both a and b are elements of the field C (or in the notation of your original post, the field D). How are you getting x4 + 2x²? That's obviously not in C², but it's not even in P2(C).

Post what you think is an example of a polynomial, together with what you think its image is under T. I'm concerned that you're trying to do this problem, but you don't know what P2(C), C², or T are.

5. Mar 15, 2006

### Mathman23

Hi I thought that I was suppose to take an abitrary polynomial
a0 + a1 + a2^2
of at most degree 2, and then insert each value of the basis B = (1,x,x^2) into that polynomial to obtain the matrix representation?

Do You then mean to say that [p(-i), p(i)] aren't roots of the polynomial??

If not what are they then?

/Fred

p.s. Is it better to look at it like a tranformation between two Vectorspaces???

Last edited: Mar 16, 2006
6. Mar 16, 2006

### HallsofIvy

You appear to be completely misunderstanding this problem.
You say
$$T: P_2(D) \rightarrow D^2$$
but then later say
"the standard base (e1,e2) in D2"
That is not a standard basis for D2- it is a standard basis for R2.
Are you sure this is not to C2 rather than D2?
I would interpret [p(-i),p(i)] to mean "apply the polynomial to -i and i (the imaginary unit) and use those values as components.

"I thought that I was suppose to take an abitrary polynomial
a0 + a1 + a2^2 of at most degree 2, and then insert each value of the basis B = (1,x,x^2) into that polynomial to obtain the matrix representation?"

Each member of the basis is a polynomial so I don't see what you mean by "insert into that polynomial".

"Do You then mean to say that [p(-i), p(i)] aren't roots of the polynomial??"

(p(-i),p(i)) are the values of the polynomial, of course, not roots. "i" here is the imaginary unit, not a variable. If p(x)= 1+ 3x+ 2x2, then p(-i)= 1+ 3(-i)+ 2(-i)2= 1- 3i- 2= -1 -3i and p(i)= 1+ 3(i)+ 2(i)2= 1+ 3i- 2= -1+ 3i.
T(p)= (

"Is it better to look at it like a tranformation between two Vectorspaces???"

Since that's what the whole problem is about, yes!

Since the domain space is three dimensional and the range is two dimensional, the linear transformation would be represented by 2 by 3 matrix, not
The simplest way to find the matrix representing a linear transformation in two given bases is to apply the transformation to each of the members of the domain basis in term and write each result in terms of the range basis. Since each basis member is represented by columns, (1, 0, 0), (0, 1, 0), (0, 0, 1), each "result" is a column in the matrix.

T(1)= (1, 1) , T(x)= (-i, i), T(x2)= (-1, -1)

Last edited by a moderator: Mar 16, 2006
7. Mar 16, 2006

### AKG

Mathman23
That's not a polynomial. [EDIT: Actually, technically it is a polynomial, but not for the reasons you think it is. It is certainly not an arbitrary polynomial, it's just a constant number]
That doesn't make any sense whatsoever.
Roots of what polynomial? p(x)? No, why would p(-i), p(i) generally be roots of the polynomial? However, I suspect you don't even understand what you mean by your own question.
It is a transformation between two vector spaces. Do you know what that means?

HallsofIvy
I think we're regarding D² as a vector space over D, so I think it makes sense to say that the standard basis would be {e1, e2}.

Last edited: Mar 16, 2006
8. Mar 16, 2006

### matt grime

But unless i is in D the question makes nosense at all, hence it is reasonble to assume i is indeed sqrt(-1) and D is C.