# Which one of the following option satisfies (1) and (2)?

Consider the subspaces $$W_1$$ and $$W_2$$ of $$\mathbb R^3$$ given by $$W_1=\{(x,y,z)\in \mathbb R^3: x+y+z=0\}$$ and $$W_2=\{(x,y,z)\in \mathbb R^3:x-y+z=0\}$$. If $$W$$ is a subspace $$\mathbb R^3$$ such that

$$(1) W\cap W_2=span\{(0,1,1)\}$$

$$(2) W\cap W_1 \text{is orthogonal to} W\cap W_2 \text{with respect to the usual innerproduct space of} \mathbb R^3.$$ Then

(A) $$W=span\{(0,1,-1),(0,1,1)\}$$

(B)$$W=span\{(1,0,-1),(0,1,-1)\}$$

(C)$$W=span\{(1,0,-1),(0,1,1)\}$$

(D)$$W=span\{(1,0,-1),(1,0,1)\}$$

Solution:- I done using the method of verification of options. I got the first option as the answer. $$W=span\{(0,1,-1),(0,1,1)\}\implies W=\{(0,x+y,-x+y):x,y \in \mathbb R \}.$$ Consider $$W \cap W_2,$$ then $$0-(x+y)+(-x+y)=0\implies x=0$$ and $$W \cap W_2=span\{(0,1,1)\}$$. Consider $$W \cap W_1,$$ then $$0+(x+y)+(-x+y)=0\implies y=0$$ and $$W \cap W_1=span\{(0,1,-1)\}$$. Hence, satisfies (1) and (2). So option (A) is the correct answer. Luckily first option is the correct answer. Else I have to verify the other options too. Is there any shortest method to solve this problem without verifying options? Please help me. This problem appeared in CSIR 2018 December.

• small question: probably you mean $W \cap W_1$ is orthogonal to $W \cap W_2$. – Student Apr 16 '19 at 15:50
• Where is the mistake? – user464147 Apr 17 '19 at 0:36
• Second condition on subsapce W. – Student Apr 17 '19 at 0:41
• Sorry! I will edit it, Than you for pointing my Typo – user464147 Apr 17 '19 at 1:01

If $$V$$ is a finite dimensional vector space and $$U,W$$ are subspaces, then $$\operatorname{dim}(U + W) + \operatorname{dim}(U \cap W) = \operatorname{dim}U + \operatorname{dim}W$$
Apply this to the subspace $$W$$ and $$W_2$$. Because $$\operatorname{dim}(W + W_2)$$ is at least $$2$$ and at most $$3$$ and $$\operatorname{dim}(W \cap W_2) = 1$$, the space $$W$$ is either one dimensional or two dimensional (so a line or a plane).
From condition (1), we know that the vector $$(0,1,1)$$ is inside $$W$$. This vector is not inside $$W_1$$, hence $$W + W_1$$ spans $$\mathbb{R}^3$$.
So assume that $$W$$ is two dimensional, in which case the statement above with the observation that $$W + W_1 = \mathbb{R}^3$$, implies that $$W \cap W_1$$ is one dimensional. It therefore suffices to find one vector orthogonal to $$(0,1,1)$$, which is inside $$W_1$$. This vector is, for example $$(0,1,-1)$$. Since this vector is inside $$W$$, we obtain that $$W$$ is the space given in answer A.
If it is not given that $$W$$ is a two dimensional space, then $$W$$ can be the line determined by the vector $$(0,1,1)$$. This space satisfies both conditions, since $$W \cap W_1$$ is the zerovector in that case (which is orthogonal to all vectors).