# Let $W_1$ and $W_2$ be two subspaces, then find the dimension of $W_1 \cap W_2$.

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.

• 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? – Minus One-Twelfth Mar 2 '19 at 3:14
• 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. – 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 $$\dim(W_1+W_2) = 2$$.
• Ok i got your idea, but can you give some hint as how to find the basis of say $W_1$? – Kushal Bhuyan Mar 2 '19 at 11:03
• Just pick literally any non-zero vector in $W_1 \setminus W_2$ and you have the second vector for a basis of $W_1$. – Robert Shore Mar 2 '19 at 16:17