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UPSC 2016 Maths Optional Paper 1 Q1b-ii — Step-by-Step Solution

3 marks · Section A

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

Question

If

W1={(x,y,z)∣x+y−z=0}W2={(x,y,z)∣3x+y−2z=0}W3={(x,y,z)∣x−7y+3z=0}\begin{aligned}W_1&=\{(x,y,z)\mid x+y-z=0\}\\ W_2&=\{(x,y,z)\mid 3x+y-2z=0\}\\ W_3&=\{(x,y,z)\mid x-7y+3z=0\}\end{aligned}

then find dim⁡(W1∩W2∩W3)\dim(W_1\cap W_2\cap W_3) and dim⁡(W1+W2)\dim(W_1+W_2).

Technique

Each plane is a 22-D subspace; dim⁡(∩)=n−rank⁡(stacked normals)\dim(\cap)=n-\operatorname{rank}(\text{stacked normals}); the sum dimension follows from dim⁡(U+V)=dim⁡U+dim⁡V−dim⁡(U∩V)\dim(U+V)=\dim U+\dim V-\dim(U\cap V).

Solution

Each WiW_i is a plane through the origin in R3\mathbb R^3, hence a 22-dimensional subspace.

Step 1 — dim⁡(W1∩W2∩W3)\dim(W_1\cap W_2\cap W_3)

The triple intersection is the solution space of the homogeneous system with coefficient matrix

M=[11−131−21−73].M=\begin{bmatrix}1&1&-1\\3&1&-2\\1&-7&3\end{bmatrix}.

Compute det⁡M\det M:

det⁡M=1(1⋅3−(−2)(−7))−1(3⋅3−(−2)⋅1)+(−1)(3⋅(−7)−1⋅1)=1(3−14)−1(9+2)−1(−22)=−11−11+22=0.\det M=1(1\cdot3-(-2)(-7))-1(3\cdot3-(-2)\cdot1)+(-1)(3\cdot(-7)-1\cdot1)=1(3-14)-1(9+2)-1(-22)=-11-11+22=0.

So rank⁡M<3\operatorname{rank}M<3. The first two rows (1,1,−1),(3,1,−2)(1,1,-1),(3,1,-2) are independent, so rank⁡M=2\operatorname{rank}M=2. Therefore

dim⁡(W1∩W2∩W3)=3−rank⁡M=3−2=1.\dim(W_1\cap W_2\cap W_3)=3-\operatorname{rank}M=3-2=1.

(Indeed W3W_3 contains the common line of W1∩W2W_1\cap W_2.)

Step 2 — dim⁡(W1+W2)\dim(W_1+W_2)

W1W_1 and W2W_2 are distinct planes (normals (1,1,−1)(1,1,-1) and (3,1,−2)(3,1,-2) are not parallel), so by the dimension formula

dim⁡(W1+W2)=dim⁡W1+dim⁡W2−dim⁡(W1∩W2)=2+2−dim⁡(W1∩W2).\dim(W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)=2+2-\dim(W_1\cap W_2).

Two distinct planes through the origin meet in a line, so dim⁡(W1∩W2)=1\dim(W_1\cap W_2)=1, giving dim⁡(W1+W2)=3\dim(W_1+W_2)=3. (Already W1+W2=R3W_1+W_2=\mathbb R^3.)

Answer

  dim⁡(W1∩W2∩W3)=1,dim⁡(W1+W2)=3.  \boxed{\;\dim(W_1\cap W_2\cap W_3)=1,\qquad \dim(W_1+W_2)=3.\;}
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