# Why these polynomials p_1, p_2, p_3, p_4 doesn't span space of degree 2 polynomial?

• Mar 30th 2011, 01:25 PM
x3bnm
Why these polynomials p_1, p_2, p_3, p_4 doesn't span space of degree 2 polynomial?
We know the definition of span:
If $S = \{v_1, v_2,....,v_r\}$ is a set of vectors in a vector space $V$, then the subspace $W$ of $V$ consisting of all linear combinations of the vectors in $S$ is called the space spanned by $v_1, v_2,....,v_r$, and we say that the vectors $v_1, v_2,....,v_r$ span $W$. To indicate that $W$ is the space spanned by the vectors in the set $S = \{v_1, v_2,....,v_r\}$, we write:

$W = span(S)$ or $W = span\{v_1, v_2, ....., v_r\}$

I've a math problem and asked to find out whether the polynomials span a space or not. Determine whether the following polynomials span $P_2$ (the set of degree 2 polynomials).

$
p_1 = 1 - x + 2.x^2, p_2 = 3 + x, p_3 = 5 - x + 4x^2, p_4 = -2 - 2x + 2x^2
$

The book says it doesn't span $P_2$

How do they figure out that $P_2$ subspace does not consist of all linear combination of polynomials $v_1, v_2,....,v_r$? How does one come to this conclusion?
• Mar 30th 2011, 02:06 PM
tonio
Quote:

Originally Posted by x3bnm
We know the definition of span:
If $S = \{v_1, v_2,....,v_r\}$ is a set of vectors in a vector space $V$, then the subspace $W$ of $V$ consisting of all linear combinations of the vectors in $S$ is called the space spanned by $v_1, v_2,....,v_r$, and we say that the vectors $v_1, v_2,....,v_r$ span $W$. To indicate that $W$ is the space spanned by the vectors in the set $S = \{v_1, v_2,....,v_r\}$, we write:

$W = span(S)$ or $W = span\{v_1, v_2, ....., v_r\}$

I've a math problem and asked to find out whether the polynomials span a space or not. Determine whether the following polynomials span $P_2$ (the set of degree 2 polynomials).

$
p_1 = 1 - x + 2.x^2, p_2 = 3 + x, p_3 = 5 - x + 4x^2, p_4 = -2 - 2x + 2x^2
$

The book says it doesn't span $P_2$

How do they figure out that $P_2$ subspace does not consist of all linear combination of polynomials $v_1, v_2,....,v_r$? How does one come to this conclusion?

You must mean whether they span or not the space of all polynomials of degree 2 or less.

Just write each given polynomial as a vector , say $1-x+2x^2\rightarrow (1,-1,2)\,,\,\,3+x\rightarrow (3,1,0)$ , etc.

Now form a 3 x 4 (or 4 x 3) matrix with the above vectors and check you get a rank 2 matrix, thus 2 vectors out of

the four are linearly independent and thus they span a 2-dimensional space, which thus

cannot be all of P_2 since this space'sdimension is 3 .

Tonio
• Mar 30th 2011, 02:07 PM
x3bnm
Never mind. I've found the answer.

$
P_2 = k_1(1 - x + 2.x^2) + k_2(3 + x) + k_3(5 - x + 4x^2) + k_4(-2 - 2x + 2x^2)
$

Set up a matrix and solve for $k_1, k_2, k_3$ and $k_4$ that's it. If you have a valid solution the vectors span and if not vectors doesn't span.
• Mar 30th 2011, 02:19 PM
x3bnm
Sorry tonio. May be we've posted at the same time. Thanks for help and explanation.