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UPSC 2021 Maths Optional Paper 1 Q2c — Step-by-Step Solution

15 marks · Section A

Subspaces · Linear Algebra · asked 6× in 13 yrs · Read the full method →

Question

Show $S=\{(x,2y,3x):x,y\text{ real}\}$ is a subspace of $\mathbb R^3(\mathbb R)$. Find two bases and the dimension.

Technique

Subspace check via parametrisation; identify spanning set; check linear independence.

Solution

Step 1 — $S$ is a subspace

Closure under +: Let $u_1=(x_1,2y_1,3x_1)$, $u_2=(x_2,2y_2,3x_2)\in S$. Then $u_1+u_2=(x_1+x_2,\;2(y_1+y_2),\;3(x_1+x_2))\in S$ (form $(x',2y',3x')$ with $x'=x_1+x_2$, $y'=y_1+y_2$).

Closure under scalar mult: $cu_1=(cx_1,2cy_1,3cx_1)\in S$.

Zero: $(0,0,0)\in S$ (take $x=y=0$).

So $S$ is a subspace ✓.

Step 2 — Parametrise

$(x,2y,3x)=x(1,0,3)+y(0,2,0)$.

So $S=\operatorname{span}\{(1,0,3),(0,2,0)\}$.

Step 3 — First basis

$B_1=\{(1,0,3),(0,2,0)\}$.

Linear independence: $\alpha(1,0,3)+\beta(0,2,0)=(0,0,0)$ gives $\alpha=0,\;2\beta=0,\;3\alpha=0$, so $\alpha=\beta=0$ ✓.

Step 4 — Second basis (rescale)

$B_2=\{(2,0,6),(0,1,0)\}$ (scale first by 2, second by 1/2).

Linear independence: similar check. Both vectors lie in $S$ (since $(2,0,6)$ has form $(x,2y,3x)$ with $x=2,y=0$; $(0,1,0)$ has form $x=0,2y=1$).

$B_2$ spans $S$ since they are scalar multiples of $B_1$ vectors.

Step 5 — Dimension

Both bases have 2 elements, so $\dim S=2$.

Answer