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 $i=1,2,3,4$.

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 $4$.

Consider
\[ W:=\{ p(x) \in P_3 \mid p(1)=0 \}. \]
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.)

Comment.

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.

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