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

10 marks · Section A

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

Question

Let $P=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\mid x-y-z=0\text{ and }2x-y+z=0\right\}$ be a collection of vectors in $\mathbb R^3(\mathbb R)$. Prove that $P$ is a subspace of $\mathbb R^3$.

Technique

Direct subspace criteria check; both defining equations are linear homogeneous, which guarantees closure under linear combinations.

Solution

Subspace criterion. A non-empty subset $P\subseteq V$ is a subspace iff

1. $0\in P$.
2. $u,v\in P\Rightarrow u+v\in P$ (closure under addition).
3. $u\in P,\,c\in\mathbb R\Rightarrow cu\in P$ (closure under scalar multiplication).

Check 1 — Zero vector

$(0,0,0)$: $0-0-0=0$ ✓ and $0-0+0=0$ ✓. So $(0,0,0)\in P$.

Hence $P$ is non-empty.

Check 2 — Closure under addition

Let $u=(x_1,y_1,z_1),\,v=(x_2,y_2,z_2)\in P$.

So $x_1-y_1-z_1=0$, $2x_1-y_1+z_1=0$, $x_2-y_2-z_2=0$, $2x_2-y_2+z_2=0$.

For $u+v=(x_1+x_2,y_1+y_2,z_1+z_2)$: $(x_1+x_2)-(y_1+y_2)-(z_1+z_2)=(x_1-y_1-z_1)+(x_2-y_2-z_2)=0+0=0$ ✓.

$2(x_1+x_2)-(y_1+y_2)+(z_1+z_2)=(2x_1-y_1+z_1)+(2x_2-y_2+z_2)=0+0=0$ ✓.

So $u+v\in P$.

Check 3 — Closure under scalar multiplication

Let $u=(x,y,z)\in P$ and $c\in\mathbb R$. Then $cu=(cx,cy,cz)$: $cx-cy-cz=c(x-y-z)=c\cdot 0=0$ ✓. $2(cx)-cy+cz=c(2x-y+z)=c\cdot 0=0$ ✓.

So $cu\in P$.

Conclusion

All three subspace criteria satisfied. $P$ is a subspace of $\mathbb R^3$. $\blacksquare$

Answer