Find basis from set of polynomials

Question

Let $P_3$ be the set of all real polynomials of degree 3 or less. This set forms a real vector space. Show that $\{2x^3+x+1,x−2,x^3−x^2\}$ is a linearly independent set, and find a basis for $P_3$ which includes these three polynomials.

---

Linear independence is easy, I just put the coefficients in a set and rref;

$\begin{bmatrix}2&1&1\\0&1&-2\\1&-1&0\end{bmatrix} => rref =>\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

So I've proven that the set is linearly independent. Also I have 3 pivots for 3 variables, so think I can write the vector space $P_3$ as a span of the polynomials (or, vectors of the coefficients);

$P_3 = span\{(2, 1, 1),(0, 1,-2),(1, -1, 0)\}$

Since a basis is simply a set of linearly independent vectors which cover the span, are these three vectors a basis for $P_3$, or do I have more work to do? (EDIT: That is, if we pretend that I hadn't completely erased an entire degree from the polynomials).

Also, given that $P_3$ is talked about in terms of polynomials (and assuming I've done the above properly); should I answer questions like this using the polynomials themselves, or is it okay to simply use vectors/matrices/etc with the coefficients?

Answer 1

Hint: how would one write the the polynomial of degree $n<1$ in your space?

The polynomial of degree $0$ is of the form $c \ x^0$. Meaning it's just a constant. This is how your matrix should look like initially. From top to bottom coefficients for $x^3,x^2,x^1,x^0$. Your vectors should have 4 components not 3. \begin{bmatrix} 2&0 &1 \\ 0&0 &-1 \\ 1&1&0 \\ 1&-2 &0 \end{bmatrix}

Oh answering your question about the polynomials or the matrices. We use the matrices because it's easier to show linear independence with them. Showing independence is equivalent to showing $A\vec x=\vec 0$ $\Rightarrow \vec x=\vec 0$. It would be hard to use polynomials themselves to show that.

The columns of the following matrix would form a basis since they are linearly independent. \begin{bmatrix} 2&0 &1 &1 \\ 0&0 &-1& 0\\ 1&1&0 & 0\\ 1&-2 &0 &0 \end{bmatrix}

Answer 2

Comment: One basis for $P_3$ is $$\{1,x,x^2,x^3\}$$ so this should tell you your set of polynomials couldn't possibly be a basis (too few elements--all bases have the same cardinality).

I have to wonder too if you really understand linear independence. If I were grading this question on a test, I would give no credit for your "proof". You should start explicitly with the hypothesis that $$a_1p_1+a_2p_2+a_3p_3=0$$ and show that this is equivalent to $$a=0,b=0,c=0$$ (here the $p_i$ are the polynomials you list).