# How to determine if $W_1\cap W_2$ is or isn't $0$, and how to then calculate the section

As an example for direct sums in my textbook they have given three vectors contained in the vectorspace $$V = \mathbb{R}^3$$:

• $$W_1 = \langle(1,0,0)^t,(0,1,0)^t\rangle$$
• $$W_1 = \langle(1,2,3)^t,(2,3,4)^t\rangle$$
• $$W_3 = \langle(1,1,1)^t\rangle$$

then $$V = W_1 + W_2$$ and $$V = W_1 + W_3$$ but $$V\neq W_2 + W_3$$ and $$V = W_1\oplus W_3$$ but $$V\neq W_1 \oplus W_2$$ because $$W_1\cap W_2 \neq 0$$

But how would I determine if the section between $$W_1$$ and $$W_2$$ is or is not equal to $$0$$ and how could i determine what that section would look like

Dropping transpositions since they don't really matter here.

Notice $$W_1$$ is simply $$\{(x,y,0)\mid x,y\in \mathbb R\}$$.

Something in $$W_2$$ looks like $$(\alpha +2\beta, 2\alpha+3\beta, 3\alpha+4\beta)$$.

Then to be in both, we must have $$3\alpha=-4\beta$$. That allows us to eliminate one of the parameters:

$$(\alpha +2\beta, 2\alpha+3\beta, 3\alpha+4\beta)=\\ (\frac{2}{3}\beta, \frac{1}{3}\beta, 0)$$.

So, the intersection is vectors of this form.

But how would I determine if the section between $$W_1$$ and $$W_2$$ is or is not equal to $$0$$

"If" is a much easier question, by the way. Since $$\dim{W_1+ W_2}=\dim{W_1}+\dim{W_2}-\dim{W_1\cap W_2}=4-\dim{W_1\cap W_2}\leq 3$$, you have that $$\dim{W_1\cap W_2}\geq 1$$, so it is nonzero.

• $3\alpha=-4\beta$ – Shubham Johri Jan 11 at 12:28
• and it should be $\dim W_1 + \dim W_2 - \dim W_1 \cap W_2$, not $-$ twice – Stockfish Jan 11 at 12:36
• @ShubhamJohri Sorry, yep :) thanks. It's corrected now – rschwieb Jan 11 at 14:12
• @Stockfish Thanks also for that typo catch! – rschwieb Jan 11 at 14:16