The standard basis for $P_2(\mathbb R)$, the vector space of quadratic polynomials of the form $ax^2+bx+c$ is the set $S=\{1,x,x^2\}$. Find bases for the subspaces of $P_2(\mathbb R)$ spanned by the following set of vectors:
a) {$-1+x-2x^2,3+3x+6x^2,9$}
b) {$1+x,x^2,-2+2x^2,-3x$}
I really have a hard time grasping basic linear algebra so I hope someone can guide me through this question?
Best Answer
The knowledge of a basis of a vector space $V$ allows us to reduce computations in $V$ to computations on matrices. If $B=\{v_1,\dots,v_n\}$ is a basis for $V$, we can define a map $C_B\colon V\to\mathbb{R}^n$ by setting $$ C_B(v)=\begin{bmatrix}\alpha_1 \\ \vdots \\ \alpha_n\end{bmatrix} \text{ if and only if } v=\alpha_1v_1+\dots+\alpha_nv_n $$ The map $C_B$ is linear and bijective, so a set $\{w_1,\dots,w_m\}$ of vectors in $V$ is linearly independent if and only if the set $$ \{C_B(w_1),\dots,C_B(w_m)\} $$ is linearly independent in $\mathbb{R}^n$. But for this we have Gaussian elimination and row echelon forms.
So, we take $B=\{1,x,x^2\}$, so that the set of vectors in $\mathbb{R}^3$ corresponding to $\{-1+x-2x^2,3+3x+6x^2,9\}$ is $$ \left\{ \begin{bmatrix}-1\\1\\-2\end{bmatrix}\,, \begin{bmatrix}3\\3\\6\end{bmatrix}\,, \begin{bmatrix}9\\0\\0\end{bmatrix}\, \right\} $$ and you just need to find the row echelon form of the matrix $$ \begin{bmatrix} -1 & 3 & 9 \\ 1 & 3 & 0 \\ -2 & 6 & 0 \end{bmatrix} $$ Without pivot reduction, it turns out to be $$ \begin{bmatrix} -1 & 3 & 9 \\ 0 & 6 & 9 \\ 0 & 0 & 18 \end{bmatrix} $$ which allows you to say that the first set is linearly independent.
The matrix to consider for case b is $$ \begin{bmatrix} 1&0&-2&0\\ 1&0&0&-3\\ 0&1&2&0 \end{bmatrix} $$ You'll easily find out that the fourth column is a linear combination of the first three, which are a linearly independent set.