UPSC 2020 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Rank and nullity; rank-nullity theorem · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Let $F$ be a subfield of complex numbers and $T$ a function from $F^3\to F^3$ defined by $T(x_1,x_2,x_3)=(x_1+x_2+3x_3,\ 2x_1-x_2,\ -3x_1+x_2-x_3)$ . What are the conditions on $a,b,c$ such that $(a,b,c)$ be in the null space of $T$ ? Find the nullity of $T$ .

Technique

Set the three output coordinates to $0$ , solve the homogeneous system; confirm with rank–nullity ( $\det=0\Rightarrow$ nontrivial kernel).

Solution

$T$ is linear (each output coordinate is a homogeneous linear form). Its matrix relative to the standard basis is

M=\begin{bmatrix}1&1&3\\ 2&-1&0\\ -3&1&-1\end{bmatrix},\qquad T\!\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=M\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.

Step 1 — Null space equations

$(a,b,c)\in\ker T \iff T(a,b,c)=(0,0,0)$ :

\begin{aligned} a+b+3c&=0 \quad (1)\\ 2a-b&=0 \quad (2)\\ -3a+b-c&=0 \quad (3) \end{aligned}

Step 2 — Solve

From (2): $b=2a$ . Substitute into (1):

a+2a+3c=0\ \Rightarrow\ 3a+3c=0\ \Rightarrow\ c=-a.

Check (3): $-3a+2a-(-a)=-3a+2a+a=0$ ✓ (automatically satisfied — the third equation is dependent).

So the conditions are

\boxed{\,b=2a\quad\text{and}\quad c=-a\,}

with $a\in F$ arbitrary. Equivalently

(a,b,c)=a(1,2,-1),\qquad a\in F.

Step 3 — Nullity

\ker T=\operatorname{span}\{(1,2,-1)\},\qquad \dim\ker T=1.

Answer

\boxed{\;\text{nullity}(T)=1.\;}