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UPSC 2014 Maths Optional Paper 1 Q2a — Step-by-Step Solution

15 marks · Section A

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

Question

Let $V$ and $W$ be the following subspaces of $\mathbb R^4$:

Find a basis and the dimension of (i) $V$, (ii) $W$, (iii) $V\cap W$.

Technique

Express subspace constraints in terms of free parameters; read off basis vectors from the parametric form.

Solution

Strategy. Solve each defining system for free parameters; express elements parametrically; read off a basis.

(i) Basis and dimension of $V$

$V$: one linear constraint $b-2c+d=0$. So $\dim V=4-1=3$.

Solve for $d=2c-b$ in terms of $a,b,c$ (the free variables). General element:

Basis of $V$: $\{(1,0,0,0),\,(0,1,0,-1),\,(0,0,1,2)\}$. $\dim V=3$.

(ii) Basis and dimension of $W$

$W$: two constraints $a=d$ and $b=2c$. So $\dim W=4-2=2$.

Solve: $a=d$, $b=2c$, with $c,d$ free. General element:

Basis of $W$: $\{(0,2,1,0),\,(1,0,0,1)\}$. $\dim W=2$.

(iii) Basis and dimension of $V\cap W$

A vector is in $V\cap W$ iff it satisfies all three constraints: $b-2c+d=0$, $a=d$, $b=2c$.

Substitute $b=2c$ into $b-2c+d=0$: $2c-2c+d=0\Rightarrow d=0$. Then $a=d=0$. Free variable: $c$.

General element: $(0,2c,c,0)=c(0,2,1,0)$.

Basis of $V\cap W$: $\{(0,2,1,0)\}$. $\dim(V\cap W)=1$.

Answer