Linear Dependent/Independent Vectors of Polynomials

Linear algebra problems and solutions

Problem 15

Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?

(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$.

(b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.

(University of California, Berkeley)

LoadingAdd to solve later


  • The set $P_3$ consisting of all polynomials of degree $3$ or less is a vector space of degree $4$.
  • Consider a subset of $P_3$ satisfying the condition (a) and prove that its dimension is less than or equal to $3$.
  • Proof.

    (a) At $1$ each of the polynomials has the value $0$.

    We show that the condition (a) is sufficient. Let us denote by $P_3$ the vector space consisting of all polynomials (with the variable $x$) of degree $3$ or less. Then the dimension of $P_3$ is $4$.

    \[ W:=\{ p(x) \in P_3 \mid p(1)=0 \}. \] Then this is a subspace of $P_3$. (Check it or see another solution below.)

    Since, for example, the polynomial $q(x)=x \in P_3$ is not in $W$, the subspace $W$ is a proper subspace of $P_3$. Hence $\dim(W)<\dim(P_3)=4$.
    (Actually, the dimension is $3$, see another solution below.)

    Since the dimension of $W$ is less than or equal to $3$, any four vectors in $W$ must be linearly dependent. Thus $p_i$ are linearly dependent.

    Another Solution of (a)

    This solution is simpler than the previous solution.

    Consider a map $P_3 \to \R$ sending $p(x)$ to $p(1)$ (evaluating at $x=1$).
    Then this is a linear transformation. The kernel of this linear transformation is
    \[ W:=\{ p(x) \in P_3 \mid p(1)=0 \}. \] (As it is the kernel, $W$ is a subspace of $P_3$.)
    By the rank-nullity theorem, we see that $\dim(W)=3$. Hence any four vectors in $W$ are linearly dependent.

    (b) At $0$ each of the polynomials has the value $1$.

    We show that the condition (b) is not sufficient.
    In fact, the following four vectors satisfy the condition (b) but they are linearly independent: $1, 1+x, 1+x+x^2, 1+x+x^2+x^3$.

    (Observe that the method of the proof of (a) does not work because
    \[V=\{ p(x) \in P_3 \mid p(0)=1 \} \] is not a subspace.)


    The second proof of (a) is simpler and I think better than the first proof.
    The reason I wrote the first proof is that it is more accessible for the first year linear algebra students.

    LoadingAdd to solve later

    More from my site

    You may also like...

    Leave a Reply

    Your email address will not be published. Required fields are marked *

    More in Linear Algebra
    Problems and solutions in Linear Algebra
    Possibilities of the Number of Solutions of a Homogeneous System of Linear Equations

    Here is a very short true or false problem. Select either True or False. Then click "Finish quiz" button. You...