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UPSC 2026 Maths Optional Paper 1 Q3a-ii — Step-by-Step Solution

7 marks · Section A

Subspaces · Linear Algebra · asked 8× in 14 yrs · Read the full method →

Question

Given $P_3[x] = \{a_0 + a_1 x + a_2 x^2 + a_3 x^3 \mid a_0, a_1, a_2, a_3 \in \mathbb{R}\}$ as a vector space over the field $\mathbb{R}$, and $H = \left\{ p(x) \in P_3[x] \;\middle|\; \displaystyle\int_{-1}^{1} p(x)\,dx = 0 \right\}$.

Find a subspace $K$ of $P_3[x]$ such that $P_3[x] = H \oplus K$.

Technique

$H$ is the kernel of the non-zero linear functional $\varphi(p)=\int_{-1}^{1}p\,dx$, and a hyperplane always has a $1$-dimensional complement: pick any $q$ with $\varphi(q)\ne 0$ and take $K=\operatorname{span}\{q\}$. Prove the direct sum by verifying the two defining conditions $H\cap K=\{0\}$ and $H+K=P_3[x]$, the second by exhibiting the explicit decomposition $p=\big(p-\frac{\varphi(p)}{\varphi(q)}q\big)+\frac{\varphi(p)}{\varphi(q)}q$.

Solution

From part (i): $\varphi(p)=\int_{-1}^{1}p\,dx$ is a linear functional on $P_3[x]$, $H=\ker\varphi$, and $\dim H = 3$ while $\dim P_3[x] = 4$.

Step 1 — Choose the candidate complement.

Take the constant polynomial $q(x)=1$. Then

Define

the set of constant polynomials. $K$ is a subspace (it is the span of a vector, hence closed under addition and scalar multiplication and contains $\mathbf{0}$), and $\dim K = 1$ since $1 \neq \mathbf 0$.

Step 2 — Prove $H \cap K = \{\mathbf 0\}$.

Let $p \in H \cap K$. Since $p \in K$, $p(x) = c$ for some $c \in \mathbb{R}$. Since $p \in H$,

Hence $p = \mathbf 0$, so

Step 3 — Prove $H + K = P_3[x]$ (constructively).

Let $p(x)=a_0+a_1x+a_2x^2+a_3x^3$ be arbitrary. From part (i), $\varphi(p)=2a_0+\frac23 a_2$. Put

and set $h = p - k$. Then $h \in P_3[x]$ and, by linearity of $\varphi$,

so $h \in H$. Explicitly,

and indeed $\int_{-1}^{1}h\,dx = 2\left(-\frac{a_2}{3}\right)+\frac23 a_2 = 0$ — consistent with the criterion $3a_0'+a_2'=0$ of part (i), since here $3\left(-\frac{a_2}{3}\right)+a_2=0$.

Therefore every $p \in P_3[x]$ is written as

so $P_3[x] = H + K$. $\qquad (\star\star)$

Step 4 — Conclude the direct sum.

By $(\star)$ and $(\star\star)$, $P_3[x]=H+K$ with $H\cap K=\{\mathbf 0\}$, which is exactly the definition of a direct sum:

Equivalently, the decomposition in Step 3 is unique: if $h_1+k_1=h_2+k_2$ with $h_i\in H, k_i\in K$, then $h_1-h_2=k_2-k_1 \in H\cap K=\{\mathbf 0\}$, forcing $h_1=h_2$ and $k_1=k_2$.

Dimension check. $\dim H + \dim K = 3+1 = 4 = \dim P_3[x]$, as a direct sum requires. Concretely, adjoining $1$ to the basis $\{1-3x^2,\,x,\,x^3\}$ of $H$ gives $\{1-3x^2,\,x,\,x^3,\,1\}$, whose coordinate matrix relative to $\{1,x,x^2,x^3\}$ is

so it is a basis of $P_3[x]$.

Step 5 — $K$ is not unique.

The argument used only $\varphi(q)\neq 0$. So for any $q\in P_3[x]$ with $\int_{-1}^{1}q\,dx\neq 0$, the line $K_q=\operatorname{span}\{q\}$ is a complement of $H$. For instance $q(x)=x^2$ gives $\varphi(x^2)=\frac23\neq 0$, so $\operatorname{span}\{x^2\}$ is another valid answer, as is $\operatorname{span}\{1+x\}$ ($\varphi = 2$). The complement $K$ is therefore not unique — the question asks only for a subspace $K$, and any one of these earns full marks provided both direct-sum conditions are proved.

Answer