# Thread: Finding the range of a Linear Transformation

1. ## Finding the range of a Linear Transformation

Hi, I'm working on the following problem and was wondering if anyone could check to see if I've solved it correctly (I can't help but wonder if my method is incorrect - it seems a little too simple).

Define $T : P_{3}(\mathbb{R}) \rightarrow P_{3}(\mathbb{R})$ by $T(f)(x) = xf^{\prime}(x)$ . Find the range of $T$

My attempt is as follows:

Let $f(x) = x^{3} + x^{2} + x + 1$ (standard form for a polynomial of $P_{3}(\mathbb{R})$)

Then $T(f)(x) = x(3x^{2} + 2x + 1) = 3x^{3} + 2x^{2} + x$

so range $T = span \{ x^{3}, x^{2}, x \}$

It seems a little crude to me for some reason...

Is it correct?

Thanks,

2. ## Re: Finding the range of a Linear Transformation

Originally Posted by anguished
Let $f(x) = x^{3} + x^{2} + x + 1$ (standard form for a polynomial of $P_{3}(\mathbb{R})$)
Could you say what a standard form for a polynomial is?

Originally Posted by anguished
Then $T(f)(x) = x(3x^{2} + 2x + 1) = 3x^{3} + 2x^{2} + x$

so range $T = span \{ x^{3}, x^{2}, x \}$
Could you explain in more detail how the statement "Range T = ..." follows?

3. ## Re: Finding the range of a Linear Transformation

Originally Posted by emakarov
Could you say what a standard form for a polynomial is?
By that I meant it's a linear combination of the elements forming the "obvious" basis for $P_{3}(\mathbb{R})$. Those elements are: $1, x, x^{2}, x^{3{$.

Originally Posted by emakarov
Could you explain in more detail how the statement "Range T = ..." follows?
The range of $T$ is all possible linear combinations of its basis elements: $x^{3}, x^{2}, x$
(These were obtained by applying the linear transformation to $f(x)$ defined above)

Does this reasoning sound right?

4. ## Re: Finding the range of a Linear Transformation

Originally Posted by anguished
By that I meant it's a linear combination of the elements forming the "obvious" basis for $P_{3}(\mathbb{R})$. Those elements are: $1, x, x^{2}, x^{3}$.
I agree that $1, x, x^{2}, x^{3}$ form the obvious bases of $P_{3}(\mathbb{R})$. However, $x^{3} + x^{2} + x + 1$ is in no way a general form or representative of all polynomials in that space; it's just a single particular polynomial. A general form would be $a_3x^{3} + a_2x^{2} + a_1x + a_0$ for some real numbers $a_3,\dots,a_0$.

Originally Posted by anguished
The range of $T$ is all possible linear combinations of its basis elements: $x^{3}, x^{2}, x$
I agree that your original conclusion "range T = span $\{ x^{3}, x^{2}, x \}$" means "The range of $T$ is all possible linear combinations of its basis elements: $x^{3}, x^{2}, x$." But how does this conclusion follow from the reasoning before that?

Originally Posted by anguished
(These were obtained by applying the linear transformation to $f(x)$ defined above)
I don't think you showed how any linear combination of $x^3, x^2, x$ except $3x^{3} + 2x^{2} + x$ was produced by T from that particular f(x).

We agree that

$\mathop{\mathrm{range}}(T)=\mathop{\mathrm{span}} \{x^3,x^2,x\}=\{a_3x^{3} + a_2x^{2} + a_1x\mid a_3,a_2,a_1\in\mathbb{R}\}=$
$\{f(x)\in P_{3}(\mathbb{R})\mid f(x)=0\}$.

A typical proof of the fact that range(T) = A for some set A consists of two parts: the first one shows range(T) ⊆ A and the second one shows A ⊆ range(T). The first part fixes an arbitrary $f\in P_{3}(\mathbb{R})$ and shows that T(f) ∈ A. The second part fixes an arbitrary g ∈ A and finds some $f\in P_{3}(\mathbb{R})$ such that T(f) = g.

5. ## Re: Finding the range of a Linear Transformation

since T(f) always has an "x" in it (unless f'(x) = 0, the 0-polynomial), one suspects that the image of T is: A = {f(x) in P3(R): f(0) = 0} (that is, polynomials with 0 constant term).

clearly T(f) is always in that set, because g(x) = x(f'(x)) always satisfies: g(0) = 0(f'(0)) = 0, no matter what f' happens to be. so we've found a containing set for T(f). T(f) ⊆ A.

now suppose g(x) = a1x + a2x2 + a3x3 = x(a1 + a2x + a3x2) is any such polynomial.

if we can find an f(x) with f'(x) = a1 + a2x + a3x2, we're done, because that's a pre-image for g.

by ordinary techniques of first-year calculus, one sees that any anti-derivative of a1 + a2x + a3x2 will do, in particular:

f(x) = a1x + (a2/2)x2 + (a3/3)x3 yields T(f) = g. so A ⊆ T(f).

geometrically, what is happening is this: differentiation "loses one dimension", so we go from P3(R) to P2(R), while "multiplication by x" acts like a shift-operator, moving the 3 coordinates we get for our derivative over one position to the right (which still leaves us with a 3-dimensional space).

6. ## Re: Finding the range of a Linear Transformation

Ah, I see. Those are more robust arguments to make. Thanks to the both of you.