Question

Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and 5x2+xâ1. The dimension of the subspace H is . Is {10x2+4xâ1,3xâ4x2+3,5x2+xâ1} a basis for P2? Be sure you can explain and justify your answer. A basis for the subspace H is { }. Enter a polynomial or a comma separated list of polynomials.

Answer 1

I suppose

$H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}$

The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$).

• Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

• Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$. Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$.

The vectors comprising $H$ form a basis for it because they are linearly independent.