UPSC 2022 Maths Optional Paper 1 Q3a-ii — Step-by-Step Solution

10 marks · Section A

Bases and dimension; coordinates in a basis · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Find a basis and dimension of $P$ , where $P$ is the subspace of $\mathbb R^3$ defined by $x-y-z=0$ and $2x-y+z=0$ .

Solve the homogeneous system for $(x,y,z)$ in terms of free parameter $z$ ; the coefficient vector of the free parameter is the basis.

x-y-z=0,\qquad 2x-y+z=0.

Subtract: $(2x-y+z)-(x-y-z)=2z\Rightarrow x+2z=0$ , i.e. $x=-2z$ .

From the first: $y=x-z=-2z-z=-3z$ .

So $(x,y,z)=(-2z,-3z,z)=z(-2,-3,1)$ .

$P=\{z(-2,-3,1):z\in\mathbb R\}=\operatorname{span}\{(-2,-3,1)\}$ .

\boxed{\;\text{Basis: }\{(-2,-3,1)\},\;\;\dim P=1.\;}