# Linear algebra - vector spaces, bases

1. Dec 8, 2015

### tawi

1. The problem statement, all variables and given/known data
1) In a vector space V of all real polynomials of third degree or less find basis B such that for arbitrary polynomial p \in V the following applies:
[p]_B = \begin{pmatrix} p'(0)\\p'(1)\\p(0)\\p(1)\end{pmatrix} where p' is the derivative of the polynomial p.

2. Relevant equations

3. The attempt at a solution
Would somebody be so kind and pointed me in a direction how to solve this? I have missed couple last lectures and I can´t seem to find this specifit type of problem anywhere.. Thanks:)

Last edited: Dec 8, 2015
2. Dec 8, 2015

### RUber

Normally, the space of third degree or less polynomials would be spanned by the standard basis:
$\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\1\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix}.$
And the polynomial
$ax^3 + bx^2+ cx + d$ would be represented by $\begin{pmatrix} d\\c\\b\\a \end{pmatrix}.$
This problem is asking you for a different way to define a basis set such that the coefficient of p in first basis will always be p'(0), the coefficient of p in the second basis will always be p'(1), and so on.
Start with what you know those values have to be for a general polynomial. Then ask yourself what combinations would give you that result...finally, test to see if you indeed have a basis.

3. Dec 8, 2015

### tawi

so for $p'(0)$ we'll have $c$, for $p'(1)$ we will have $3a^2+2b+c$, for $p(0) d$ and for $p(1) a^3+b^2+c+d$... How will the final solution look like then? Just a vector with these elements?

4. Dec 8, 2015

### epenguin

Have you made an error in transcription of the problem? Maybe there should be a p(1) or something?

5. Dec 8, 2015

### tawi

Yes, the last one should in fact be p(1).

6. Dec 8, 2015

### epenguin

Would the question in ordinary algebra be: express any polynomial ax3 + bx2 + cx + d in terms of those other things, two of which are already c and d.
That is not very difficult in ordinary algebra, so maybe you could express your result back into vector space terms, and maybe the ordinary calculation back into vector space terms.
Not being very handy with vector spaces, I would like to see it.

7. Dec 8, 2015

### tawi

You mean writing $ax^3 + bx^2 + cx + d = p$ which will give us system of four equations with the same LHS and RHS will always be the result for the individual elements of the basis?
So the first equation would be $ax^3 + bx^2 + cx + d = c$, second $ax^3 + bx^2 + cx + d = d$, third $ax^3 + bx^2 + cx + d = 3a^2 + 2b + c$ and $ax^3 + bx^2 + cx + d = a^3 + b^2 + c+ d$?

8. Dec 8, 2015

### epenguin

No I don't think I meant that, that doesn't seem to me to mean anything.
I did mean just obtain a, b, c, d in terms of the elements you were given.

I don't know what your conventions are for writing polynomials - RUber seems to be leaving out x's altogether.
My concrete way would be a horizontal vector
(x0, x1, x2,x3) that multiplies a matrix made of RUber's columns that multiplies his vertical vector.
I could change that into a different matrix that multiplies the vector in the problem.

9. Dec 8, 2015

### RUber

Note that p'(1) will not have coefficients raised to powers. The x terms were the variables raised to powers and you are plugging in x=1 to the polynomial.

Example p(1)= a(1^3)+b(1^2)+c(1)+d= a+b+c+d.