does the set of vectors span P3?

I'm having trouble understanding this homework problem.

Suppose four polynomials are defined by the following:

$p_{1}(x) = x^3 - 2x^2 + x + 1 \\ p_{2}(x) = x^2 - x + 2 \\ p_{3}(x) = 2x^3 + 3x + 4 \\ p_{4}(x) = 3x^2 + 2x + 1 \\$

Does the set $S =$ { ${p_{1}, p_{2}, p_{3}, p_{4}}$ } span $P_{3}$ (the space of all polynomials of degree at most 3)?

So if I start with the polynomial $y = ax^3 + bx^2 + cx + d$ I understand(I think) that in order for $S$ to span $P_{3}$ then $y$ must be a linear combination of $S$ but I'm not sure where to go from there.

EDIT:

$$A = \begin{bmatrix} 1 & -2 & 1 & 1\\ 0 & 1 & -1 & 2\\ 2 & 0 & 3 & 4\\ 0 & 3 & 2 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

So $rank(A) = 4$ which means the vectors in the set are linearly independent because there are only 4 column vectors in the matrix (AND there is only the trivial solution to the matrix) and therefore we span $P_{3}$?

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One way might be to note that since $S$ contains 4 vectors and $P_3$ has dimension 4, then $S$ spans $P_3$ iff the vectors in $S$ are linearly independent. –  Tobias Kildetoft Feb 1 '12 at 22:03
hint: write the matrix of the coefficients of the polynomials, and calculate its rank –  Emilio Ferrucci Feb 1 '12 at 22:06

Since $P_3$ has dimension 4, the set $\{p_1,p_2,p_3,p_4\}$ will span $P_3$ if and only if they are independent. So, test if the vectors in $\{p_1,p_2,p_3,p_4\}$ are independent:

Assume $$c_1p_1+c_2p_2+c_3p_3+c_4p_4={\bf 0}.$$ Then $$c_1(x^3-2x^2+x+1)+c_2(x^2-x+2)+c_3(2x^3+3x+4)+c_4(3x^2+2x+1) ={\bf 0}.$$ Collecting like terms, the above can be written as $$(c_1+2c_3)x^3+(-2c_1+c_2+3c_4)x^2+(c_1-c_2+3c_3+2c_4)x+(c_1+2c_2+4c_3+c_4)={\bf 0}.$$ A polynomial is the zero polynomial if and only if all its coefficients are 0; so, the above is equivalent to the following system of equations: \tag{1}\eqalign{ c_1+ 2c_3 &=0\cr -2c_1+c_2+3c_4&=0\cr c_1-c_2+3c_3+2c_4&=0\cr c_1+2c_2+4c_3+c_4&=0} The coefficient matrix of the above system is $$A=\left[\matrix{1&0&2&0\cr -2&1&0&3\cr 1&-1&3&2\cr 1&2&4&1\cr }\right]$$ An echelon form of $A$ is $$\left[\matrix{1&0&2&0\cr 0&1&-4&3\cr 0&0&-3&5\cr 0&0&0&7\cr }\right].$$ This implies the system above has only the trivial solution: $c_1=c_2=c_3=c_4=0$; thus $\{p_1,p_2,p_3,p_4\}$ is an independent set and so spans $P_3$.

Alternatively you could show that the equation you wrote always has a solution. To do this, write the corresponding system of equations. You'll wind up with the system as in $(1)$, but with the right hand side replace by the coefficients of $y$.

If you just want to "cut to the chase", note that the coefficient matrix $A$ above is simply the matrix whose columns are the coefficients of the polynomials $p_1$, $p_2$, $p_3$, $p_4$. You could have immediately written this down (or written them down as rows) and then determined if the matrix has full rank. It's a good thing to see why you can do this, though...

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This was exactly what I was attempting to do. I just couldn't make the connection from my polynomials to the matrix. Thank you. –  intervade Feb 1 '12 at 23:08

Hint: The answer can be shortened by determining the determinant of $A$ and see if it is not equal to $0$. Hence by Theorem if the determinant is not equal to zero the set is linearly independent else its linearly dependent.

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Which Theorem do you mean? Clarify it please. –  f.nasim May 28 '13 at 19:29