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

8 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 $\mathbb{R}$.

Let $H = \left\{ p(x) \in P_3[x] \;\middle|\; \displaystyle\int_{-1}^{1} p(x)\,dx = 0 \right\}$. Prove that $H$ is a subspace of $P_3[x]$ and find a basis of $H$.

Technique

Recognise $H$ as the kernel of the linear functional $\varphi(p)=\int_{-1}^{1}p\,dx$ on $P_3[x]$: linearity of the integral gives the subspace property in three lines, and the rank–nullity theorem pins $\dim H = 4-1 = 3$ before any basis is written down. Then convert the single scalar condition into a relation on the coefficients and read off a basis.

Solution

Step 1 — Set up the defining map and show it is linear.

Define $\varphi : P_3[x] \to \mathbb{R}$ by

For $p,q \in P_3[x]$ and $\alpha,\beta \in \mathbb{R}$, the integral is linear:

So $\varphi$ is a linear functional on $P_3[x]$, and by definition

Step 2 — $H$ is a subspace.

We verify the subspace criterion directly (not merely by citing “kernels are subspaces”).

(i) $H \neq \varnothing$; the zero vector lies in $H$. The zero polynomial $\mathbf{0}$ belongs to $P_3[x]$ and $\int_{-1}^{1} 0\,dx = 0$, so $\mathbf{0} \in H$.

(ii) Closure under addition. Let $p, q \in H$. Then $p+q \in P_3[x]$ (a sum of polynomials of degree $\le 3$ has degree $\le 3$), and

so $p+q \in H$.

(iii) Closure under scalar multiplication. Let $p \in H$, $\alpha \in \mathbb{R}$. Then $\alpha p \in P_3[x]$ and

so $\alpha p \in H$.

Being a non-empty subset of $P_3[x]$ closed under addition and scalar multiplication, $H$ is a subspace of $P_3[x]$. $\blacksquare$

Step 3 — Translate the condition into the coefficients.

For $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$,

The odd powers $x$ and $x^3$ integrate to $0$ over the symmetric interval $[-1,1]$, leaving

Hence

with $a_1, a_3$ completely free.

Step 4 — Dimension by rank–nullity.

$\varphi$ is not the zero functional, since $\varphi(1) = \int_{-1}^{1}1\,dx = 2 \neq 0$. Therefore $\operatorname{Im}\varphi$ is a non-zero subspace of the $1$-dimensional space $\mathbb{R}$, so $\operatorname{rank}\varphi = 1$. By rank–nullity applied to $\varphi : P_3[x] \to \mathbb{R}$ with $\dim P_3[x] = 4$:

So any spanning set of $H$ with three elements is automatically a basis.

Step 5 — Exhibit a basis.

Using $a_2 = -3a_0$, a general element of $H$ is

with $a_0, a_1, a_3 \in \mathbb{R}$ arbitrary. Hence

Independence. Suppose $\alpha(1-3x^2) + \beta x + \gamma x^3 = 0$ as a polynomial, i.e.

Equating coefficients of $1, x, x^2, x^3$ (the standard basis of $P_3[x]$) gives $\alpha = 0$, $\beta = 0$, $-3\alpha = 0$, $\gamma = 0$. So $\alpha=\beta=\gamma=0$ and the three polynomials are linearly independent.

They are three independent vectors spanning the $3$-dimensional space $H$, so they form a basis. (Each does satisfy the defining condition: $\int_{-1}^{1}(1-3x^2)dx = 2 - 3\cdot\frac23 = 0$, and $\int_{-1}^{1}x\,dx = \int_{-1}^{1}x^3dx = 0$ by oddness.)

Remark (an equally valid, more structural basis). The Legendre polynomials $P_1(x)=x$, $P_2(x)=\frac{3x^2-1}{2}$, $P_3(x)=\frac{5x^3-3x}{2}$ are each orthogonal to $P_0 = 1$ on $[-1,1]$, i.e. $\int_{-1}^{1}P_n\,dx = 0$ for $n \ge 1$. Being non-zero polynomials of distinct degrees $1,2,3$ they are independent, so $\{P_1,P_2,P_3\}$ is also a basis of $H$. A basis of $H$ is of course not unique.

Answer