Basis of vector space of all polynomials degree 3 and lower. 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.
 A: 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.
A: You've learned lots of techniques for answering such questions about matrices — so turn the question into a problem about matrices. Express everything in coordinates relative to some basis (e.g. the power basis $1,x,x^2,x^3$), and note things like:


*

*$ g \mapsto g(3)$ is a linear transformation on $W$; in coordinates, it is multiplying on the right by the matrix
$$ \left[ \begin{matrix}1 \\ 3 \\ 3^2 \\ 3^3 \end{matrix} \right] $$

*The coordinates of the elements of $V_1$ are the rows of the matrix, where I've used the symbol $\cdot$ for a matrix element whose value you don't actually care about.
$$ \left[ \begin{matrix} \cdot & 1 & 0 & 0
\\ \cdot & \cdot & 1 & 0
\\ \cdot & \cdot & \cdot & 1 \end{matrix} \right]$$


(In the above, I used the convention of writing the coordinates of the elements of $W$ as row vectors. Using column vectors for this tends to be much more natural — however, you're probably more familiar with studying matrices in terms of row operations, and it's that part of the analysis that's important here, so I've used the opposite convention instead)
