0
$\begingroup$

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.

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

Supposing you know the following statement:

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy