# Finding a basis for a set of polynomials and functions

trap101
Find a basis for and the dimension of the subspaces defined for each of the following sets of conditions:

{p $\in$ P3(R) | p(2) = p(-1) = 0 }

{ f$\in$Span{ex, e2x, e3x} | f(0) = f'(0) = 0}

Attempt: Having trouble getting started...

So I think my issue is interpreting what those sets are and setting it up. So I think the sets are: i) the set of all polynomials s.t P(2) = p(-1) = 0 and ii) the set of exp functions where at 0 equal 0.

So how do I put these each into a matrix form to find the basis and dimension?

Homework Helper
Gold Member
Find a basis for and the dimension of the subspaces defined for each of the following sets of conditions:

{p $\in$ P3(R) | p(2) = p(-1) = 0 }

{ f$\in$Span{ex, e2x, e3x} | f(0) = f'(0) = 0}

Attempt: Having trouble getting started...

So I think my issue is interpreting what those sets are and setting it up. So I think the sets are: i) the set of all polynomials s.t P(2) = p(-1) = 0
I doubt this. It is probably the set of all polynomials with degree <= 3, such that p(2) = p(-1) = 0.
and ii) the set of exp functions where at 0 equal 0.
Not just any exp functions. They have to be in $\text{span}(e^x, e^{2x}, e^{3x})$.

I suggest that you start by finding the dimension of these two spaces: $P_3(\mathbb{R})$ and $\text{span}(e^x, e^{2x}, e^{3x})$. Also, what is the form of a general element for each of these two spaces?

Homework Helper
Personally, I wouldn't use a matrix, I would use the basic definition. First, I am going to assume that P3 is the vector space of polynomials of degree 3 or less, which has dimension 4 (some texts use that to mean the space of polynomials 2 or less which has dimension 3- the same ideas will apply but it is simpler). Any such polynomial canbe written $p(x)= ax^3+ bx^2+ cx+ d$. The condition that p(2)= 0 means that we must have $p(2)= 8a+ 4b+ 2c+ d= 0$ or $d= -(8a+ 4b+ 2c)$. The condition that p(-1)= 0 means that $-a+ b- 2c+ d= 0[/tex] or $$d= -(a- b+ 2c)$$. Then d= -(8a+ 4b+ 2c)= -(a- b+ 2c) so that 8a+ 4b+ 2c= a- b+ 2c which reduces to 7a= -5b. Sp we can replace a by -5b/7 which means d= -(-(5/7)b- b+ 2c)= -(12/7)b- 2c. Using those, [itex]ax^3+ bx^2+ cx+ d= -(5/7)bx^3+ bx^2+ cx- (12/7)b- 2c= (-(5/7)x^3+ x^2- 12/7)b+ c(x- 2)$

Now, do you see what a basis is and what the dimension is?
(You could have made a quick "guess" at what the dimension is by the fact that the basic space has dimension 4 and there are 2 conditions put on it.)

For the second one, any f in the span of ex, e2x, and e3x, can be written as f(x)= aex+ be2x+ ce3x, and f'(x)= aex+ 2bex+ 3cex.

The condition that f(0)= 0 gives a+ b+ c= 0 and f'(0)= 0 gives a+ 2b+ 3c= 0. We can subtract the first equations from the second to get b+ 2c= 0 or b= -2c. Putting that into the first equation a- 2c+ c= a- c= 0 so a= c. That is, we can write aex+ bex+ cex= cex[/sup- 2ce2x+ ce3x= c(ex- 2e2x+ e3x. Now, what is the dimension and what is a basis?
(Here, the basic space has dimension three and there are two conditions.)

trap101
I doubt this. It is probably the set of all polynomials with degree <= 3, such that p(2) = p(-1) = 0.

Not just any exp functions. They have to be in $\text{span}(e^x, e^{2x}, e^{3x})$.

I suggest that you start by finding the dimension of these two spaces: $P_3(\mathbb{R})$ and $\text{span}(e^x, e^{2x}, e^{3x})$. Also, what is the form of a general element for each of these two spaces?

That's what I intend on doing, but my issue is setting it up in order ot find those dimensions. So here's how I'm trying to piece it together:

I know the general form for P3(R) is: ax3+bx2+cx+d, now the condition is that p(2) = P(-1) = 0. So I have to some how write out a set of vectors that satisfy that form.

As for ii) a function would be f(x) = ex-2e2x+e3x, but I'm utterly clueless as to how this is line independent and how I could even find this vector if I set up a matrix