Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$T:P_3\rightarrow \mathbb{R}$ 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.

share|improve this question
6  
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
show 1 more comment

2 Answers 2

up vote 1 down vote accepted

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 $\ldots$

share|improve this answer
    
"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
add comment

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

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 $\,\dim V-1\,$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.