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UPSC Maths 2013 Paper 1 Q2c-ii — Solution

8 marks · Section A

Question

Show that the vectors $X_1=(1,1+i,i),\,X_2=(i,-i,1-i)$ and $X_3=(0,1-2i,2-i)$ in $\mathbb{C}^{3}$ are linearly independent over the field of real numbers but are linearly dependent over the field of complex numbers.

Technique

For complex dependence, solve the linear system; for real independence, exploit that a real scalar times $i$ has zero real part — so the equation $a+bi=0$ with $a,b\in\mathbb{R}$ forces both to be zero.

Solution

Part 1 — Dependence over $\mathbb{C}$

Strategy. Find complex scalars $a,b,c$ (not all zero) with $aX_1+bX_2+cX_3=0$. Try $c=1$ and solve for $a,b$.

Suppose $aX_1+bX_2=X_3$ (i.e., $c=-1$). Componentwise:

• Coord 1: $a+bi=0$.
• Coord 2: $a(1+i)+b(-i)=1-2i$.
• Coord 3: $ai+b(1-i)=2-i$.

From coord 1: $a=-bi$.

Substitute into coord 2: $-bi(1+i)-bi=1-2i\;\Rightarrow\;-bi-bi^{2}-bi=1-2i\;\Rightarrow\;b-2bi=1-2i\;\Rightarrow\;b(1-2i)=1-2i\;\Rightarrow\;b=1$.

Then $a=-bi=-i$.

Check coord 3: $ai+b(1-i)=(-i)(i)+1\cdot(1-i)=1+1-i=2-i$ ✓.

So $X_3=-iX_1+X_2$, equivalently

The coefficients $(i,-1,1)$ are complex (not all zero), so $\{X_1,X_2,X_3\}$ is linearly dependent over $\mathbb{C}$.

Part 2 — Independence over $\mathbb{R}$

Strategy. Show that the only real solution to $aX_1+bX_2+cX_3=0$ is $a=b=c=0$.

Let $a,b,c\in\mathbb{R}$ satisfy

Coord 1: $a\cdot 1+b\cdot i+c\cdot 0=a+bi=0$.

Since $a,b\in\mathbb{R}$, the real and imaginary parts must vanish separately: $a=0$ and $b=0$.

Coord 3: substitute $a=b=0$: $0+0+c(2-i)=0$. Since $2-i\ne 0$, $c=0$.

So $a=b=c=0$.

Answer