The vectors in
form a basis of
if they are mutually linearly independent and span
.
To check for independence, we can compute the Wronskian determinant:
![\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7Dx%5E2%2B3x%2B1%262x%5E2%2Bx-1%264%5C%5C2x%2B3%264x%2B1%260%5C%5C2%264%260%5Cend%7Bvmatrix%7D%3D4%5Cbegin%7Bvmatrix%7D2x%2B3%264x%2B1%5C%5C2%264%5Cend%7Bvmatrix%7D%3D40%5Cneq0)
The determinant is non-zero, so the vectors are indeed independent.
To check if they span
, you need to show that any vector in
can be expressed as a linear combination of the vectors in
. We can write an arbitrary vector in
as
![p=ax^2+bx+c](https://tex.z-dn.net/?f=p%3Dax%5E2%2Bbx%2Bc)
Then we need to show that there is always some choice of scalars
such that
![k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p](https://tex.z-dn.net/?f=k_1%28x%5E2%2B3x%2B1%29%2Bk_2%282x%5E2%2Bx-1%29%2Bk_34%3Dp)
This is equivalent to solving
![(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c](https://tex.z-dn.net/?f=%28k_1%2B2k_2%29x%5E2%2B%283k_1%2Bk_2%29x%2B%28k_1-k_2%2B4k_3%29%3Dax%5E2%2Bbx%2Bc)
or the system (in matrix form)
![\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%261%260%5C%5C3%261%260%5C%5C1%26-1%264%5Cend%7Bbmatrix%7D%5Cbegin%7Bbmatrix%7Dk_1%5C%5Ck_2%5C%5Ck_3%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7Da%5C%5Cb%5C%5Cc%5Cend%7Bbmatrix%7D)
This has a solution if the coefficient matrix on the left is invertible. It is, because
![\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7D1%261%260%5C%5C3%261%260%5C%5C1%26-1%264%5Cend%7Bvmatrix%7D%3D4%5Cbegin%7Bvmatrix%7D1%262%5C%5C3%261%5Cend%7Bvmatrix%7D%3D-20%5Cneq0)
(that is, the coefficient matrix is not singular, so an inverse exists)
Compute the inverse any way you like; you should get
![\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D1%261%260%5C%5C3%261%260%5C%5C1%26-1%264%5Cend%7Bbmatrix%7D%5E%7B-1%7D%3D-%5Cdfrac1%7B20%7D%5Cbegin%7Bbmatrix%7D4%26-8%260%5C%5C-12%264%260%5C%5C-4%263%26-5%5Cend%7Bbmatrix%7D)
Then
![\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Dk_1%5C%5Ck_2%5C%5Ck_3%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D1%261%260%5C%5C3%261%260%5C%5C1%26-1%264%5Cend%7Bbmatrix%7D%5E%7B-1%7D%5Cbegin%7Bbmatrix%7Da%5C%5Cb%5C%5Cc%5Cend%7Bbmatrix%7D)
![\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}](https://tex.z-dn.net/?f=%5Cimplies%20k_1%3D%5Cdfrac%7B2b-a%7D5%2Ck_2%3D%5Cdfrac%7B3a-b%7D5%2Ck_3%3D%5Cdfrac%7B4a-3b%2B5c%7D%7B20%7D)
A solution exists for any choice of
, so the vectors in
indeed span
.
The vectors in
are independent and span
, so
forms a basis of
.