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.

I suppose[tex]H=\mathrm{span}\{10x^2+4x-1,3x-4x^2+3,5x^2+x-1\}[/tex]The vectors that span [tex]H[/tex] form a basis for [tex]P_2[/tex] if they are (1) linearly independent and (2) any vector in [tex]P_2[/tex] can be expressed as a linear combination of those vectors (i.e. they span [tex]P_2[/tex]).Independence:Compute the Wronskian determinant:[tex]\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[/tex]The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of [tex]H[/tex] is 3.Span:Write an arbitrary vector in [tex]P_2[/tex] as [tex]ax^2+bx+c[/tex]. Then the given vectors span [tex]P_2[/tex] if there is always a choice of scalars [tex]k_1,k_2,k_3[/tex] such that[tex]k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c[/tex]which is equivalent to the system[tex]\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}[/tex]The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives[tex]\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}[/tex]so the vectors do span [tex]P_2[/tex].The vectors comprising [tex]H[/tex] form a basis for it because they are linearly independent.