# Linear Transformation from P2 (R) to P3 (R)

## Homework Statement

Let T be the linear transformation from P2 (R) to P3 (R) defined by
$$T(f)=14\int_{0}^{x}f(t)dt + 7x.f'(x)$$
for each
$$f(x)=ax^{2}+bx+c$$

Determine a basis {g1, g2, g3} for Im(T).

as above

## The Attempt at a Solution

I evaluated the transformation function and simplified it, and found 3 linearly independent vectors for the basis, but I'm only asked for 1 and I think I've made a blunder somewhere.
$$\int_{0}^{x}f(t)dt=\frac{1}{3}ax^{3}+\frac{1}{2}bx^{2}+cx$$
$$f'(x)=2ax+b$$
$$T(f)=\frac{14}{3}ax^{3}+7bx^{2}+14cx+14ax^{2}+7bx$$
$$T(f)=\frac{14}{3}ax^{3}+(14a+7b)x^{2} + (7b+14c)x$$
$$T(f)= a(\frac{14}{3}, 14, 0)+b(0,7,7)+c(0,0,7)$$
$$\therefore {(\frac{14}{3}, 14, 0),(0,7,7),(0,0,7)}$$ is a basis for Im(T)

I'd be pleased if someone could eyeball it for me and see if I haven't done something really silly!

## Answers and Replies

I'm assuming by $P_n(\mathbf{R})$ you mean the set of polynomials with real coefficients with degree less than or equal to $n$; i.e. $P_n(\mathbf{R})=\{p\in\mathbf{R}[x]:\text{deg}(p)\leq n\}$.

So, first things first. It looks like you may have a misconception (albeit a reasonable and common misconception) about what a general vector space is. A vector space is not necessarily a set of $n$-tuples or arrows. In other words, the elements of a vector space need not be what we would normally call vectors. I'll direct you to Wikipedia for the full definition (http://en.wikipedia.org/wiki/Vector_space#Definition).

In your case, $P_n(\mathbf{R})$ is an $\mathbf{R}$-vector space. You can add two polynomials in $P_n(\mathbf{R})$ and get another polynomial in $P_n(\mathbf{R})$. You can multiply any polynomial in $P_n(\mathbf{R})$ by a real number to get another polynomial in $P_n(\mathbf{R})$. This addition and multiplication follow all of the rules for vector spaces. Note that polynomials are not arrows or $n$-tuples (though there is an "obvious" way to think of them as tuples, and they're often defined formally as a particular type of infinite tuples). The dimension of $P_n(\mathbf{R})$ is $n+1$.

If we're not thinking of polynomials as $n$-tuples (and I don't think we should here if we want to get the most out of this exercise), then a basis for $P_n(\mathbf{R})$ is a set of polynomials that satisfies all of the requirements for being a basis (http://en.wikipedia.org/wiki/Basis_(linear_algebra)#Definition, pay no attention to the picture). For instance the set (not tuple) $\{1,x\}$ is a basis for $P_1(\mathbf{R})$. Most would consider it to be the canonical basis for $P_1(\mathbf{R})$. $\{1,1+x\}$ is also a basis for $P_1(\mathbf{R})$, as is $\{x,1+x\}$. Can you verify that the sets I've given are in fact bases? If we were to think of polynomials with degree less than or equal to 1 as tuples, what 2-tuples would we associate with the polynomials $1$, $x$, and $1+x$?

Can you give the canonical basis for $P_2(\mathbf{R})$? Can you give another? Check that it's a basis.

Alright. Now that we have that out of the way. The fun thing is that we can apply most (if not all) of the theorems from Linear Algebra to our finite vector spaces $P_n(\mathbf{R})$. In particular, we can say the following:

If $\{g_1, g_2, ... , g_{n+1}\}$ is a basis for $P_n(\mathbf{R})$ and $T:P_n(\mathbf{R})\rightarrow P_m(\mathbf{R})$ is a linear map, then $\{T(g_1), T(g_2), ... , T(g_{n+1})\}$ is a basis (though not necessarily linearly independent) for $T(P_n(\mathbf{R}))=Im(T)$.

Is this enough to get you going?

P.S. It would probably be a good idea to prove (if only to yourself) that the transformation $T$ given in your problem is, in fact, linear.

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Thanks for your informative reply gopher_p. I think I have most of these concepts down, although my language is lacking. What I meant by 'vector' was actually 'column vector', 'tuple' is probably a less ambiguous word.

To answer your questions about P1(R). To verify that the sets you have given are indeed bases, I need to check that the elements in the set are linearly independent (ie. one is a not scalar multiple of another) and that the set is a spanning set (ie. I can generate every vector in the P1(R) space from a linear combination of each element in the set)

To verify that {1, 1 + x} is a basis, firstly it must be linearly independent because there is no way to get 1 + x from any multiple of 1, and vice versa. Secondly, to check that it is a spanning set, I need to show that I can generate any polynomial in P1(R) from this basis, by using a linear combination of each element:

a(1) + b(1 + x) = ux + v

I can see intuitively that this is so, but that's not the same as proving it. I will come back to this.

The canonical basis for P2(R) is {(0,1) , (1,0)}, and likewise for Pn(R) it would be {(1, 0, .., 0), (0, 1, ..., 0) ..., (0, 0, ..., 1)} - each tuple is all 0 except for a 1 in the nth position.

I would like to answer the questions of yours that I've skipped, but I'm going to study some more first, because it's clear that I'm missing a few things.

Thanks again, I'll be back!

The canonical basis for P2(R) is {(0,1) , (1,0)}, and likewise for Pn(R) it would be {(1, 0, .., 0), (0, 1, ..., 0) ..., (0, 0, ..., 1)} - each tuple is all 0 except for a 1 in the nth position.

What elements of $P_2(\mathbf{R})$ do $(0,1)$ and $(1,0)$ represent? How can two elements possibly form a basis for a vector space of dimension 3?

I strongly encourage you to proceed with this problem and with your understanding of vector spaces without making reference to tuples or vectors (column or row).

Oh wow, sorry I don't know what I was thinking there. The canonical basis for P2(R) - that of polynomials with degree less than or equal to 2 - is {1, x, x^2}, I gave the basis for R2 and Rn.

In any case, back to my book.