Given $W_1=\{(u,v,w,x)\in \mathbb{R^4}: u+v+w=0, 2v+x=0, 2u+2w-x=0\}$ and $W_2=\{(u,v,w,x)\in \mathbb{R^4}: u+w+x=0, u+w-2x=0, v-x=0\}$, then which of the following is true?

  1. dim $W_1=1$

  2. dim $W_2=2$

  3. dim$W_1\cap W_2=1$

  4. dim $W_1+W_2=3$

For $W_1$, there are only two linearly independent restrictions, because $u+v+w=\frac{1}{2}(2v+x+2u+2w-x)$. So dim$W_1=2.$

Again, for $W_2,$ all the restrictions are linearly independent, hence dim$W_2=1$.

Now all I have to determine the dim$W_1\cap W_2.$ For that I need the bases of both $W_1$ and $W_2$. I am facing problem here. How to find the basis by taking all the restrictions into account? Can anybody give me a hint? Thanks.

  • $\begingroup$ Hint: one possibility to find the desired dimension (not necessarily the best way) is to write down the set of conditions that $\mathbf{x}:= \begin{bmatrix}u\\ v\\ w\\ x\end{bmatrix}\in \mathbb{R}^4$ must satisfy to be in $W_1 \cap W_2$ in the form $A\mathbf{x} = \mathbf{0}$ for some particular matrix $A$. Then $W_1 \cap W_2 = \left\{ \mathbf{x} \in \mathbb{R}^4 : A\mathbf{x} = \mathbf{0}\right\}$. Do you know how to find the dimension of a space like this? $\endgroup$ – Minus One-Twelfth Mar 2 '19 at 3:14
  • $\begingroup$ Also a hint for one way to find a basis for $W_1$ if you want to ($W_2$ is similar) is to do similar to above: write $W_1$ as a set of conditions matrix form as above, and use row reduction to find a basis. $\endgroup$ – Minus One-Twelfth Mar 2 '19 at 3:17

There are only two possibilities. Either $W_2 \subseteq W_1$ or $W_2 \cap W_1 = \{0 \}$. And in this case, it's not hard to see which vectors are in $W_2$. Adding the first two defining equations, we see that $-x=0$, so $x=0$. The third equation then tells us $v=0$, and in light of those two facts, the first (and second) equation tells us $u=-w$. So $W_2 = \{(t, 0, -t, 0)~|~t \in \Bbb R \}$.

Vectors of this form always solve the equations defining $W_1$ so $W_2 \subseteq W_1$. Thus, $\dim(W_1 \cap W_2) = 1$ and $\dim(W_1+W_2) = 2$.

  • $\begingroup$ Ok i got your idea, but can you give some hint as how to find the basis of say $W_1$? $\endgroup$ – Kushal Bhuyan Mar 2 '19 at 11:03
  • $\begingroup$ Just pick literally any non-zero vector in $W_1 \setminus W_2$ and you have the second vector for a basis of $W_1$. $\endgroup$ – Robert Shore Mar 2 '19 at 16:17

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