Question: Find two different bases for the vector space $W$ of all third and lower degree polynomials $g(x)$ satisfying $g(3)=0$.
Don't make the basis simply scalar multiples of the first basis.
Here is my solution
$$V_1 = \{(x-3),(x-3)^2,(x-3)^3\}\quad \quad
V_2 = \{(3-x),x(3-x),x^2(3-x)\}$$
I am pretty sure that $V_1,V_2$ both span the polynomials with degree $3$ and lower, how should I go about showing that. Also how do I show $V_1,V_2$ are not scalar multiples of each other.
Best Answer
Yes remove 1 from both the sets. This being a proper subspace, its dimension cannot be 4. Its dimension is 3. You can directly calculate and see that $a(x-3)^2\neq x(3-x)$ for any $a$ etc. (Assume such an equality is true and show that you cannot solve for $a$. This will lead to an inconsistent linear system. Same way for the next.
Also your second basis can be proved to be a basis this way:
Fact: In polynomials with real coefficients the operation of multiplying them all by a fixed polynomial is an injective function: $f(x)g(x)=f(x)h(x)$ implies $g(x)=h(x)$ (with obvious exemptions). ALso this operation is linear. That means the linearly independent set of polynomials $\{1,x,x^2\}$ upon multiplication by $(3-x)$ will again be sent to a linearly independent set, and that is your second basis.