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UPSC 2026 Maths Optional Paper 1 Q4a-i — Step-by-Step Solution

8 marks · Section A

Rank and nullity; rank-nullity theorem · Linear Algebra · asked 9× in 14 yrs · Read the full method →

Question

Let $F$ be a subfield of complex numbers and let $T$ be the linear transformation from $F^3$ into $F^3$, defined by

If $(a, b, c)$ is a vector in $F^3$, then answer the following:

Find the condition on $a, b, c$ such that the vector $(a, b, c)$ be in the null space of $T$. What is the nullity of $T$?

Technique

Write $T$ as a matrix $A$ in the standard basis; membership of the null space is the homogeneous system $A(a,b,c)^{T}=0$. Row-reduce once — the rref simultaneously delivers the explicit condition on $a,b,c$ and the nullity $3-\operatorname{rank}A$. Because $A$ has integer entries, the reduction is valid over every subfield $F\subseteq\mathbb{C}$.

Solution

Step 1 — Matrix of $T$ in the standard basis.

With $\mathcal{B}=\{e_1,e_2,e_3\}$ the standard ordered basis of $F^3$,

so, writing the images as columns,

Step 2 — Set up the defining condition.

By definition

that is, the homogeneous system

Step 3 — Row-reduce the coefficient matrix.

Continuing to reduced form ($R_2\to\frac13 R_2$, then $R_1\to R_1+R_2$):

There are two non-zero rows, so

(Consistency check on the determinant: $\det A = 1(1\cdot2-0\cdot(-2))+1(2\cdot2-0\cdot(-1))+2(2\cdot(-2)-1\cdot(-1)) = 2+4-6 = 0$, so $A$ is singular and $\operatorname{rank}A\le 2$; rows $1,2$ are non-proportional, so $\operatorname{rank}A=2$ exactly.)

Step 4 — Read off the condition on $a,b,c$.

The rref gives $a+\frac23 c=0$ and $b-\frac43 c=0$, i.e.

Equivalently — and this is the tidier pair to quote, because it uses the given equations untouched — work directly from $(1),(2),(3)$. From $(2)$: $b=-2a$. Substituting in $(1)$: $a+2a+2c=0$, i.e. $3a+2c=0$. Equation $(3)$ then reads $-a-2(-2a)+2c=3a+2c=0$, which is the same equation — so $(3)$ carries no new information.

The dependence is exact and worth recording: with $R_1=(1,-1,2)$, $R_2=(2,1,0)$, $R_3=(-1,-2,2)$,

so $R_3=R_1-R_2$: the third equation is the first minus the second. Exactly one of the three equations is dependent, confirming $\operatorname{rank}A=2$.

Hence the condition is the pair of equations

equivalently, in proportional (symmetric) form,

Step 5 — The null space and the nullity.

One variable is free (nullity $=3-\operatorname{rank}=1$). Put $a=2t$; then $b=-2a=-4t$ and $c=-\tfrac32 a=-3t$, so

Hence

a one-dimensional subspace (the spanning vector is non-zero, so it is a basis), and

Direct check on the basis vector:

Step 6 — Rank–nullity cross-check, and independence of the field.

Finally, note that $A$ has integer entries and the entire row reduction used only the rational scalars $-2,\,1,\,\frac13$. Any subfield $F$ of $\mathbb{C}$ has characteristic $0$ and therefore contains $\mathbb{Q}$; so all these operations are legitimate in $F$, and the rref — hence the rank, the nullity, and the condition on $a,b,c$ — is the same for every subfield $F\subseteq\mathbb{C}$. The answer does not depend on the choice of $F$.

Answer