# Linear Transformations on Function Spaces

So I'm working on homework for my Introductory Linear Algebra Course. The text is Gareth Williams Linear Algebra with Application - 7th Edition. I'm currently working section 4.7: Kernel, Range and the Rank/Nullity Theorem.

Question: Problem 20. Prove that $T: P_3\rightarrow P_2$ defined as followed in linear. Find the Kernel and the Range of T. Give bases for these subspaces. $$T(a_3x^3 + a_2x^2 + a_1x + a_0) = a_3x^2 - a_0$$ I'm fairly comfortable with proving that the above is linear, that is I can show that the transformation preserves the operations of addition and scalar multiplication. I show this by taking 2 polynomials in $P_1$ and $P_2$ and showing that $T(P_1 + P_2)$ = $T(P_1) + T(P_2)$ and that $T(cP_1)$ (where $c$ is an element of the Reals) $= cT(P_1)$

Now I understand that the kernel is the subset that maps to 0, which I believe in our case should be the zero polynomial. I'm stuck as to how about finding the kernel. I know that the range shares the dimension of the range and that the dimension of the range is equal to the rank. What I don't know how to do is to deal with this using polynomials. I'm comfortable with vectors that I can use to build a transition matrix, I just don't know how to work this through with polynomials. I feel like I could comfortably find a basis again with vector or matrix spaces but transferring these ideas to polynomials and function I struggle with. Help?

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If you enclose your math type in dollar signs (or dollar for a separate line), it'll look a lot prettier and easier to read. –  andybenji Dec 2 '12 at 22:25
Little comment on notation: you should probably use lower-case $p$'s for your polynomials and reserve the upper case $P$'s for your polynomial spaces. Otherwise you'll run in to confusions like "let $P_3\in P_3$". –  icurays1 Dec 2 '12 at 23:39

The kernel of this map is the following: $$ker(T) = \{ a_0 + a_1 x + a_2 x^2 + a_3 x^3 \in P_3 : a_3 x^2 - a_0 = 0\}$$

If this polynomial $a_3 x^2 - a_0$ is equal to zero for all $x$ values, then we know that $a_3$ and $a_0$ must be zero. So the kernel is in fact the set of all polynomials in $P_3$ with $a_3 = a_0 = 0$, in other words, all polynomials of the form $a_1 x + a_2 x^2$.

You can somewhat check this result using rank nullity. The dimension of $P_3$ is equal to $4$. The dimension of the image is $2$, and hence the dimension of the kernel is $4-2 = 2$.

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So would my basis for the kernel then be {x, x^2} and my basis for the range be {1, x^3} This seems to fall into line with the rank - nullity theorem? –  Ben Anderson Dec 2 '12 at 23:22
That is correct! –  andybenji Dec 3 '12 at 3:44

To find the kernel of an operator, it's usually a good idea to start with an object in the range of the operator, then find out how that object can be zero, then figure out where it needed to come from. So, for all $p\in P_3$ we know that

$$T(p)=a_3x^2-a_0$$

So the range looks like quadratic polynomials of the form $\alpha x^2-\beta$. How can such a polynomial be zero? Well, "zero" in this case means the zero polynomial, which is the polynomial that has all zero coefficients. Thus for $\alpha x^2-\beta$ to be the zero polynomial, we must have $\alpha=\beta=0$. So, $T(p)=0\Leftrightarrow a_3=a_0=0$. What about $a_1$ and $a_2$? They can be anything!

Sometimes when working with finite dimensional function spaces like polynomials, its easier to fix a basis and then just work with the coefficients of the polynomial under that basis. For instance, if we fix the basis $\{1,x,x^2,x^3\}$ for $P_3$, we can represent a polynomial $a_3x^3+a_2x^2+a_1x+a_0$ by its vector of coefficients, $(a_0,a_1,a_2,a_3)^\intercal$. Then you can represent operators like your $T$ as matrices and go from there (bonus question, what is the matrix of your $T$ if you use the basis I menioned?)

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