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

7 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 range of $T$. What is the rank of $T$?

Technique

$(a,b,c)\in R(T)$ exactly when the non-homogeneous system $A(x,y,z)^T=(a,b,c)^T$ is consistent. So row-reduce the augmented matrix $[A\,|\,(a,b,c)]$ and force the last row to be non-contradictory: the resulting single linear equation in $a,b,c$ is the required condition. The rank is then read off the reduced coefficient block, and cross-checked against $\operatorname{nullity}=1$ from part (i) by rank–nullity.

Solution

Step 1 — Matrix form and what “in the range” means.

As in part (i), in the standard basis

By definition,

i.e. iff the system

is consistent. By the Rouché–Capelli (consistency) criterion this happens iff

Step 2 — Row-reduce the augmented matrix.

Step 3 — Read off the consistency condition.

The last row asserts $0\cdot x+0\cdot y+0\cdot z = b+c-a$. This is satisfiable iff the right-hand side vanishes. Hence

If $a=b+c$ the system is consistent (with one free variable, matching the nullity $1$); if $a\ne b+c$ the third row reads $0=$ (non-zero), and $(a,b,c)$ is not in the range.

Step 4 — The rank of $T$.

The reduced coefficient block has exactly two non-zero rows, so

Independently: $R(T)$ is spanned by the images of the basis vectors, i.e. by the columns of $A$,

Each column satisfies $a-b-c=0$:

The first two columns are not scalar multiples of each other, so they are independent and $\operatorname{rank}T\ge 2$. The third column is dependent on them: solving $\alpha(1,2,-1)+\beta(-1,1,-2)=(2,0,2)$, the second coordinate gives $2\alpha+\beta=0$ and the first gives $\alpha-\beta=2$, whence $3\alpha=2$, $\alpha=\tfrac23$, $\beta=-\tfrac43$; the third coordinate then checks: $-\tfrac23-2\left(-\tfrac43\right)=-\tfrac23+\tfrac83=2$. $\checkmark$ So

and $\operatorname{rank}T=2$ exactly.

Step 5 — The range as a subspace, and the rank–nullity cross-check.

a $2$-dimensional subspace (over $F=\mathbb{R}$, the plane $x-y-z=0$ through the origin). Both spanning vectors are genuinely attained:

With part (i),

which is the rank–nullity theorem. $\checkmark$

Step 6 — Independence of the field $F$.

Every row operation used above involved only the integers $-2,1$ and the rational $\tfrac13$. Any subfield $F\subseteq\mathbb{C}$ has characteristic $0$, hence contains $\mathbb{Q}$, so these operations are all available in $F$. Therefore the condition $a=b+c$ and the value $\operatorname{rank}T=2$ hold for every subfield $F$ of $\mathbb{C}$ — the answer is field-independent.

Remark (a free extra structural fact). The null-space generator $(2,-4,-3)$ has $a-b-c=2+4+3=9\ne 0$, so $(2,-4,-3)\notin R(T)$. Hence $N(T)\cap R(T)=\{0\}$ and, dimensions adding to $3$,

Answer