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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?

user115636
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  • Are you familiar with row reduction technique? – alex Mar 01 '14 at 14:50
  • yes I read about it but I'm not sure how to make this into echelon form without making it really complicated?? @alex – user115636 Mar 01 '14 at 14:55
  • It is quite simple but a bit lengthy to explain. Can't you find a step-by-step explanation already written down? Even on Wikipedia. Anyway in both cases the vectors are independent, so the subspace they span is the whole vector space. You see this when you row-reduce the matrices with their components in the rows and observe that no row is identically zero. – alex Mar 01 '14 at 15:02

2 Answers2

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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.

egreg
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I'll do for (a), you do for (b): suppose

$$a\cdot(-1+x-2x^2)+b\cdot (3+3x+6x^2)+c\cdot9=0\;,\;\;a,b,c\in\Bbb R\implies$$

$$=-a+3b+9c+\left(a+3b\right)x+\left(-2a+b\right)x^2$$

Since you know $\;\{1,x,x^2\}\;$ is a basis, the above trivial linear combination is possible iff

$$\begin{cases}-a&+3b+9c&=&0\\{}\\\;\;a&+3b&=&0\\{}\\\!\!\!-2a&+b&=&0\end{cases}$$

Solving the above linear system, one gets:

$$2II+III=0\implies7b=0\implies b=0\implies a=0\implies c=0$$

and thus the given set's lin. independent.

DonAntonio
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