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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

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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.

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

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