Linear dependency of polynomials question

Solution 1:

You need to know whether there are scalars $c_1$, $c_2$, $c_3$ not all $0$ such that $c_1p_1+c_2p_2+c_3p_3=0$. So: $$ \begin{align} & c_1p_1(x)+c_2p_2(x)+c_3p_3(x) \\[6pt] = {} & c_1 (2-x^2)+c_2(3x)+c_3(x^2+x-2) \\[6pt] = {} & (c_3-c_1)x^2 + (3c_2 + c_3)x + 2(c_1-c_3) \\[6pt] = {} & 0 \text{ for all }x. \end{align} $$ That means you need $$ \begin{align} c_3-c_1& =0 \\[6pt] 3c_2 + c_3 & =0 \\[6pt] 2(c_1-c_3) & =0 \end{align} $$ The question is whether than can happen if at least one of the numbers $c_1,c_2,c_3$ is not $0$. And the answer is "yes", as you should be able to figure out from there. Therefore they are linearly dependent.

Three members of a $3$-dimensional vector space cannot span the space unless they're linearly independent. (More than three can, in some cases.)

Solution 2:

$\{1,x,x^2\}$ form a basis for the space of quadratic polynomials, and you can represent your polynomials as vectors in terms of this basis. Therefore you simply need to determine whether the matrix $$\left(\begin{matrix}-1 & 0 & 2\\0 & 3&0\\1&1&-2\end{matrix}\right)$$is invertible. Plainly it is not.

Solution 3:

Assuming that $P^{(2)}$ represents all polynomials over some field of degree less than or equal to $2$, then your questions are the same, because the dimension of $P^{(2)}$ is $3$, so if the given polynomials are linearly independent, they must form a basis of $P^{(2)}$ and hence span it.

To answer the question of linear independence, we can associate each polynomial to a vector by taking it's leading coefficients. The vectors obtained in this way are $(-1,0,2)$, $(0,3,0)$, and $(1,1,-2)$. To see that these vectors are linearly independent, compute the determinant of the matrix they determine: $$\begin{pmatrix} -1&0&2\\ 0&3&0\\ 1&1&-2 \end{pmatrix}$$

It turns out that the determinant of this matrix is zero, so the vectors are not linearly independent. A dependence relation is given by:

$$3p_1(x)-p_2(x)+3p_3(x)=0$$

Solution 4:

We could do it "by inspection." Add the first and the third.

So we have $3$ vectors, not independent. They cannot span a $3$-dimensional space.