# Does this set of vectors span the space of all polynomials of degree at most 3?

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 $\operatorname{rank}(A) = 4$ 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
Although it is not important for computing the rank, your matrix recording the polynomials is transpose to what it normally should be: coordinates of individual vectors should define columns of the matrix. The advantage of doing so is that should the row reduction have revealed the rank to be less than $4$, then it would also have provided you explicitly with the linear relation(s) between those vectors (namely the solutions of the homogeneous system with the RREF as left hand side gives such relations). – Marc van Leeuwen Jan 30 '15 at 6:33

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
Actually computing the determinant is not shorter than row reducing a matrix. – Marc van Leeuwen Jan 30 '15 at 7:27