# Thread: Is polynomial of degree 3 linearly independant

1. ## Is polynomial of degree 3 linearly independant

All,

Let V be the vector space of all polynomials of degree 3 at most. Is the set

$\displaystyle A=\left({x^3+2x+3, x^2+4x-1, x+5,2x}\right)$ linearly independent? Is A a basis for V

First of all i want to establish the difference between 'basis' and 'its dimension'. I am thinking they mean the same thing...ie, $\displaystyle {\mathbb{R}}^3$ is of dimension 3 because you need 3 linearly independant vectors to get to a point in 3D space. THis is the same as saying it has a basis involving 3 vectors.....correct me if im wrong...

Thanks

2. ## Re: Is polynomial of degree 3 linearly independant

Originally Posted by bugatti79
All,

Let V be the vector space of all polynomials of degree 3 at most. Is the set

$\displaystyle A=\left({x^3+2x+3, x^2+4x-1, x+5,2x}\right)$ linearly independent? Is A a basis for V

First of all i want to establish the difference between 'basis' and 'its dimension'. I am thinking they mean the same thing...ie, $\displaystyle {\mathbb{R}}^3$ is of dimension 3 because you need 3 linearly independant vectors to get to a point in 3D space. THis is the same as saying it has a basis involving 3 vectors.....correct me if im wrong...

Thanks
You are correct that the dimention of a space tells you how many linearly independant vectors are required to span a space.

In your example of $\displaystyle \mathbb{R}^3$ it says that the basis must have 3 vectors in it

One basis could be

$\displaystyle v_1=(1,0,0) \quad v_2=(0,1,0) \quad v_3=(0,0,1)$ these are linearly independant and span $\displaystyle \mathbb{R}^3$

but you could also have the basis

$\displaystyle v_1=(1,0,0) \quad v_2=(1,1,0) \quad v_3=(1,1,1)$ these are also linearly independant and span $\displaystyle \mathbb{R}^3$

but they both still have 3 vectors.

One way to solve the above problem is to express the polynomials in terms of the standard basis of $\displaystyle P_{3}$

The basis is $\displaystyle \{1, x, x^2, x^3 \}$ So for example the vector $\displaystyle 2x^2-1=(-1,0,2,0)$. Now the vectors may look more familiar!

3. ## Re: Is polynomial of degree 3 linearly independant

Originally Posted by TheEmptySet
The basis is $\displaystyle \{1, x, x^2, x^3 \}$
Before I attempt to solve this.

Is the above basis a dimension of 4? Ie hypothetically in 4D space we need these 4 elements to define a point in 4D space?
IF so...how did you know what to include in the basis? ie I would have left out the '1' but for no valid reason....

Thanks

4. ## Re: Is polynomial of degree 3 linearly independant

one of the things a basis must be is a spanning set. note than any polynomial of degree 3 or less can be written:

$\displaystyle p(x) = a_0 + a_1x + a_2x^2 + a_3x^4$, so $\displaystyle \{1,x,x^2,x^3\}$ is a spanning set, and it should be clear

that if p(x) = 0 for all x, that all 4 coefficients $\displaystyle a_0,a_1,a_2,a_3$ must be 0, so we have linear independence as well (the other thing a basis must be).

(if we left out the element 1 from our basis, we couldn't have any constant polynomials).

5. ## Re: Is polynomial of degree 3 linearly independant

Originally Posted by TheEmptySet
You are correct that the dimention of a space tells you how many linearly independant vectors are required to span a space.

In your example of $\displaystyle \mathbb{R}^3$ it says that the basis must have 3 vectors in it
Where does it say the basis must have 3 vectors in it? The degree 3 is referring to the polynomial...

The definition of a basis is that the set (x_1,x_2....x_n) is a finite basis for a vector space if

a) it is linearly independent which we have established
b) the set spans V ie span(x_1,x_2...x_n)

How is b) shown mathematically?

THanks

6. ## Re: Is polynomial of degree 3 linearly independant

in general, to prove that span(x1,...,xn) = V, you have to show that any v in V can be written:

v = a1x1+...+anxn,

that is you have to show you can find "coordinates" a1,...,an that make this true.

in the case of R3, we can show {(1,0,0),(0,1,0),(0,0,1)} spans R3 rather easily:

(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1).

the same logic shows {1,x,x2,x3} spans P3(R), given a polynomial of degree 3 of less,

say p(x) = a0 + a1x + a2x2 + a3x3,

we see that p(x) is indeed a linear combination (in the span-set) of 1,x,x2 and x3

(these play the role of the standard basis in R4, to which P3(R) is isomorphic).

in fact, we have a "change-of-basis" linear transformation given by:

p(x) = a0 + a1x + a2x2 + a3x3 ↔ (a0,a1,a2,a3),

which maps "basis to basis":

1 ↔ (1,0,0,0)
x ↔ (0,1,0,0) (you can think of this as 0(1) + 1(x) + 0(x2) + 0(x3))
x2 ↔ (0,0,1,0)
x3 ↔ (0,0,0,1)