# Vector Space: Basis

1. May 8, 2005

### laminatedevildoll

In the vector space P_4 of all polynomials of degree less than or equal to 4 we define the first five Tchebychev polynomial as

p_0(x) = 1
p_1(x) = x
p_2(x) = 2x^2 - 1
p_3(x) = 4x^3 - 3x
p_4(x) = 8x^4 - 8x^2 + 1

To show that B={p_0, p_1, p_2, p_3, p_4} is a basis of P_4, do I put them in a matrix, find the row-echelon form and find the vector space with pivots? I so, how do I put it in a matrix?

2. May 8, 2005

### Galileo

To use a matrix, you need to choose a basis first. You know that {1,x,x^2,x^3,x^4} forms a basis for P_4 so you can use that.

I think it's easier to use the definition of linear independence, suppose:

$$ap_0(x)+bp_1(x)+cp_2(x)+dp_3(x)+ep_4(x)=0$$
for all x.

You can write it out and see that you must have a=b=c=d=e=0, or use the different degrees of the polynomials.

3. May 8, 2005

### HallsofIvy

Setting up a matrix having the coefficients as entries is the same as using the "standard" basis: 1, x, x2, x3, x4. Do that and "row reduce". If you do not get one or more rows that are all 0, then the four functions are independent and so form a basis for P4.

Or course, Galileo is also correct. Write out the 4 equations, take some simple values like x= 0, x= 1, x= -1, x= 2, for example to get 4 equations for the four coefficients a, b, c, d and solve. If they must all be 0, then the functions are independent and form a basis.

4. May 8, 2005

### matt grime

Or you could just observe that they are trivially linearly independent, and hence must be a basis.

5. May 8, 2005

### Galileo

I used to give such answers on my exams, but I never did get points for that...

6. May 8, 2005

### matt grime

exams should never have questions for which the answer genuinely is trivial, though. at least they don't if i set them. A better question to ponder might be:

suppose the p_0,...,p_n is a family of polynomials, and that the degre of p_r is r. Show that these are a basis of the space of all polys of degree at most n if and only if p_0 is not zero.

7. May 8, 2005

### mathwonk

all you have to do is glance at them to see they are a basis. i.e. each ahs a different degree. isn't that enough?

8. May 9, 2005

### matt grime

yep, which is why i labelled the problem trivial, in the reasonably mathematical sense, but at least this way you can't plug in numbers to get simultaneous questions to solve.

9. May 9, 2005

### mathwonk

On "triviality" in mathematics:

I once handed in a hw problem in abstract algebra to find the multiplicative inverse of the non zero complex number a+bi, by simply saying it was [a-bi]/[a^2+b^2],
since this problem was one i had solved years before, and that I by then considered it rather trivial.

the grader replied "no points for copying the answer from the back of the book."

I was incensed, at his implication of both dishonesty and stupidity on my part, as until that time I had never been aware of the practice of some authors of putting answers in the back of books. After learning it, I did not care to look anyway, having more confidence in my own answers than those of some paid drudge.

But I lost all respect for the moron grader at that point, and stopped even handing in hw for his approval.

To this day, visitors to my office occasionally notice that the answer book for the calculus course I teach is used as a heavy doorstop.

In graduate school, I once answered a hw question on sheaf cohomology as follows:

Q: show that if you have a short exact sequence of sheaves, then the corresponding sequence of global sections is left exact.

A: the easy result that the kernel of a map of presheaves is in fact a sheaf implies this problem.

At this point the grader stopped reading in despair, but I had in fact followed this true assertion by a complete explanation of the solution.

As best I can recall what I assumed the grader knew was that exactness as presheaves implies exactness as sheaves, hence if the presheaf kernel is a sheaf, then it is also the sheaf kernel, whence it follows from definition of presheaf exactness that the sequence of global sections is left exact.

I had actually given a two page explanation of this result but the grader did not even read it after my comment as to the triviality of the problem. In my defense, I was merely imitating the writing style of the most popular authors of that time, and also, of many of our lecturers.

Today however it seems politically incorrect to inform people that something is trivial, so in my own writings I try to avoid use of the words trivial or obvious, (except on this site, late at night.) In reality, to be told something is trivial, is a lot of information.

A co author and I once asked some colleagues for the written version of their announced results so we could use them in our own paper, but they merely responded that it had not yet appeared and was "easy" anyway. Armed with that knowledge, we sat down and proved it ourselves, at which point they quickly provided their version.

So when someone reveals that a problem is trivial, you should look quickly to see why. If I had read Matt's post I should not have offered mine, but I merely saw with some amazement that a trivial question was getting a lot of ink, and tried to dispatch it as mercifully as possible.

by the way if you have learned or memorized something about echelon form, note the matrix of these coefficients is already in echelon form, and none of the rows is zero, so you are done.

Last edited: May 9, 2005