understanding basics of linear transformations, in polynomials

$T:P_3\rightarrow \mathbb{R}$ given: $T(p)= \int_0^1x^2p(x)dx$. Prove that T is a linear transformation, and find a basis for its kernel.

DO NOT SOLVE

My textbook explain using only very abstract terms, and as I'm new to the concept; I'm having difficulty in applying the idea of 'linear transformations'.

Questions:

what is the term $P_3$... it is not defined. I'm guessing a polynomial of at most 3rd degree. Further is $p(x)$ then also a polynomial of 3rd degree?

Any hints/ advise would be much appreciated.

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I was sure that one of these days, one of my students would show up here posting a question from one of my assignments. I hope that whatever help you get here you credit when you turn your work in. Better would have been to ask for help from within. –  Gerry Myerson Sep 29 '12 at 5:53
ill rephrase question, to only clarification...unless u still think its too much –  student101 Sep 29 '12 at 5:59
Professors are on this site, but I guess you've just figured that out. That being said, the thing you're transforming is $p$, not $x$, so subbing in $p=ax+by$ is incorrect. Separate out the transformation from the thing you're transforming. That should make it clearer. –  Robert Mastragostino Sep 29 '12 at 6:00
i see thankyou. that makes more sense –  student101 Sep 29 '12 at 6:01
We have used the notation $P_n$ for the vector space of all polynomials of degree at most $n$ --- I'm sure you'll find it in the notes you've taken at lectures. The first clause tells you the domain of $T$ is $P_3$, so, yes, when you see $T(p)$, $p$ is meant to be an element of $P_3$. Not necessarily a polynomial of degree 3, but of degree at most 3. Have you read the notes I put up at rutherglen.science.mq.edu.au/math133s212/notes/… –  Gerry Myerson Sep 29 '12 at 6:59

I assume $P_3$ is the set of polynomials with real coefficients having degree $\leq3$.

How does the general element $p\in P_3$ look like? (Note that a polynomial, albeit being a complicated expression, is considered here as a "point" in $P_3$ and is denoted therefore by a single letter $p$.)

Verify: $P_3$ is a vector space.

Produce a basis of $P_3$, i.e., an array $\bigl(e_k)_{1\leq k\leq n}$ of special polynomials such that any $p\in P_3$ can be written as a linear combination (with real coefficients) of these $e_k$. What is $n$ ?

Verify that $T:\ P_3\to{\mathbb R}$ is linear, i.e., that for arbitrary $p$, $q\in P_3$ and arbitrary $\alpha$,$\beta\in{\mathbb R}$ one has $T(\alpha p+\beta q)=\alpha T(p)+\beta T(q)$.

Now we are proceeding to the analysis of $T$. To do this we have to find the matrix of $T$. How many rows resp. columns does this matrix have? To find the matrix we have to compute the effect of $T$ on the $n$ basis vectors and to write the result into the columns of the matrix. This means that we actually have to compute $n$ integrals.

This should do for the moment $\ldots$

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"To do this we have to find the matrix of $T$." As it happens, the students haven't seen the idea of the matrix representing a linear transformation. The problem can be done without finding the matrix of $T$. –  Gerry Myerson Sep 30 '12 at 7:52

$$P_3:=\{f(x)\in\Bbb R[x]\;;\;\deg f\leq 3\}$$

Thus , $\,p(x)\,$ is a real polynomial of degree at most $\,3\,$.

One last point/hint: $\,T\,$ is a linear transformation from a vector space to its definition field (which is a vector space of dimension $\,1\,$ over itself) , so in fact it's what is also called a linear functional, and these have the very nice following property:

Lemma: Any non-zero linear functional on any vector space $\,V\,$ is onto and thus its kernel is a hyperplane in $\,V\,$ , i.e.: a maximal proper subspace of $\,V\,$, which in finite dimension ammounts to be of dimension $\,\dim V-1\,$

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