Linear dependency of polynomials question

Question

I have to determine whether the polynomials $p_1(x)=2-x^2$, $p_2(x)=3x$, $p_3(x)= x^2 +x-2$ are linearly dependent or independent but I am not sure how to start. Anyone care to enlighten me?

Also I have to find out if it spans $P^{(2)}$.

Answer 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.)

Answer 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.

Answer 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:

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

Answer 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.

Answer 5

Consider the problem this way: By definition, the linear independence of terms in a polynomial function is to show that $$ k_0+k_1\cdot x+k_2\cdot x^2 +\cdots+k_n\cdot x^n=0. \tag{1} $$ only if $k_0, k_1, k_2, ..., k_n$ are all zero. To show this is the case, take derivatives of $(1)$ successively up to the $n$th derivative; then we can form a system of homogeneous linear equations for the $k$s. The coefficient matrix of the system of equations is $$ \begin{bmatrix} 1& x& x^2& \cdots& x^{n-1}& x^n\\ 0& 1& 2x& \cdots& (n-1)x^{n-2}& nx^{n-1}\\ 0& 0& 2& \cdots& (n-1)(n-2)x^{n-3}& n(n-1)x^{n-2}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\\ 0& 0& 0& \cdots& 0& n!\\ \end{bmatrix} $$ The determinant of the above matrix is a product of all the diagonal elements, which is 2!3!...(n-1)!n!, obviously not zero regardless of $x$. Therefore the only solutions for $k$s are all zero. This proves the independence.