Exercise 1923Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

A subspace of polynomials with a root condition

#polynomial subspace#evaluation#dimension

Statement

Let $F=\{P\in\mathbb{R}_3[X] : P(1)=0\}$ . Show that $F$ is a subspace of $\mathbb{R}_3[X]$ and compute its dimension.

Hint

The evaluation map $P\mapsto P(1)$ is a linear form. It is surjective. Use rank-nullity.

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Linearity. The map $\varphi:P\mapsto P(1)$ is linear (evaluation is linear). So $F=\ker\varphi$ is a subspace.

Surjectivity. $\varphi$ is nonzero (e.g. $\varphi(1)=1$ ), hence surjective onto $\mathbb{R}$ .

Dimension. $\dim F=\dim\mathbb{R}_3[X]-1=4-1=3$ .

Explicit basis. $P\in F$ iff $P$ is divisible by $X-1$ , i.e. $P=(X-1)Q$ with $Q\in\mathbb{R}_2[X]$ . A basis is $(X-1, (X-1)X, (X-1)X^2)$ .

Conclusion. $F$ is a hyperplane of $\mathbb{R}_3[X]$ , of dimension $3$ .