UPSC Maths 2025 Paper 1 Q4c-ii — Solution

8 marks · Section A

Question

Let $P_n$ denote the vector space of all polynomials of degree $\leq n$ over $\mathbb{R}$ . Verify that $\dim\!\left(\dfrac{P_4}{P_2}\right) = \dim P_4 - \dim P_2$ .

Technique

Find the dimensions of the polynomial spaces from the monomial basis, then verify the quotient-dimension formula $\dim(V/W)=\dim V-\dim W$ by exhibiting an explicit basis of the quotient.

Solution

Step 0 — $P_2$ is a subspace of $P_4$ . Every polynomial of degree $\le2$ also has degree $\le4$ , so $P_2\subseteq P_4$ , and it is closed under addition and scalar multiplication. Hence $P_2$ is a subspace of $P_4$ and the quotient $P_4/P_2$ is well-defined.

Step 1 — Dimensions of $P_4$ and $P_2$ .

A basis of $P_n$ is $\{1,x,x^2,\dots,x^n\}$ , so $\dim P_n=n+1$ . Therefore $\dim P_4=5\quad(\text{basis }\{1,x,x^2,x^3,x^4\}),\qquad \dim P_2=3\quad(\text{basis }\{1,x,x^2\}).$ Hence $\dim P_4-\dim P_2=5-3=2.$

Step 2 — Dimension of the quotient $P_4/P_2$ directly.

Cosets in $P_4/P_2$ are $p(x)+P_2$ . Two polynomials are in the same coset iff their difference lies in $P_2$ , i.e. iff they have the same degree-3 and degree-4 coefficients. Writing a general element of $P_4$ as $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4,$ its coset is $p(x)+P_2=a_3\,(x^3+P_2)+a_4\,(x^4+P_2),$ because $a_0+a_1x+a_2x^2\in P_2$ collapses to the zero coset. Thus every coset is a unique linear combination of the two cosets $\overline{x^3}=x^3+P_2$ and $\overline{x^4}=x^4+P_2$ .

Independence of $\{\overline{x^3},\overline{x^4}\}$ : if $\alpha\overline{x^3}+\beta\overline{x^4}=\overline 0$ , then $\alpha x^3+\beta x^4\in P_2$ , forcing $\alpha=\beta=0$ (a nonzero degree-3 or degree-4 polynomial cannot have degree $\le2$ ). So $\{\overline{x^3},\overline{x^4}\}$ is a basis of $P_4/P_2$ and $\dim\!\left(\frac{P_4}{P_2}\right)=2.$

Step 3 — Compare.

$\dim\!\left(\frac{P_4}{P_2}\right)=2=5-3=\dim P_4-\dim P_2.\qquad\blacksquare$

This is the special case of the general theorem $\dim(V/W)=\dim V-\dim W$ (which follows from the rank–nullity theorem applied to the quotient map $\pi:V\to V/W$ , whose kernel is $W$ ).

Answer

$\boxed{\;\dim\!\left(\frac{P_4}{P_2}\right)=2=\dim P_4-\dim P_2=5-3.\;}$