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UPSC 2017 Maths Optional Paper 1 Q2d — Step-by-Step Solution

10 marks · Section A

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

Question

Suppose $U$ and $W$ are distinct four dimensional subspaces of a vector space $V$, where $\dim V=6$. Find the possible dimensions of subspace $U\cap W$.

Technique

$\dim(U\cap W)=\dim U+\dim W-\dim(U+W)$, then bound $\dim(U+W)$ between $\max(\dim U,\dim W)=4$ and $\dim V=6$, using distinctness to exclude the top end.

Solution

Step 1 — Dimension formula

For finite-dimensional subspaces,

Here $\dim U=\dim W=4$, so

Step 2 — Bound $\dim(U+W)$ from above

$U+W$ is a subspace of $V$, so $\dim(U+W)\le\dim V=6$. Hence

Step 3 — Bound $\dim(U\cap W)$ from above

$U\cap W\subseteq U$, so $\dim(U\cap W)\le\dim U=4$. But $U$ and $W$ are distinct, so $U\cap W$ cannot equal both; if $\dim(U\cap W)=4$ then $U\cap W=U=W$ (a $4$-dim subspace contained in a $4$-dim space is the whole space), contradicting distinctness. Thus $\dim(U\cap W)\le 3$.

Step 4 — Both values are attainable

Combining, $2\le\dim(U\cap W)\le 3$, so the only candidates are $\mathbf{2}$ and $\mathbf 3$. Both occur:

• $\dim=3$: in $\mathbb R^6=\langle e_1,\dots,e_6\rangle$ take $U=\langle e_1,e_2,e_3,e_4\rangle$, $W=\langle e_1,e_2,e_3,e_5\rangle$. Then $U\cap W=\langle e_1,e_2,e_3\rangle$, dimension $3$, and $\dim(U+W)=5\le6$.
• $\dim=2$: take $U=\langle e_1,e_2,e_3,e_4\rangle$, $W=\langle e_1,e_2,e_5,e_6\rangle$. Then $U\cap W=\langle e_1,e_2\rangle$, dimension $2$, and $\dim(U+W)=6$.

Answer