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I have the following 2 subspaces:

$F_1 = \{(x, y, z) \in \mathbb{R}^3, x+y=0\}$

$F_2 = \{(x, y, z) \in \mathbb{R}^3, x+z=y\}$

Given that, we have:

$F_1 = \{(x_1(1, -1, 0) + z_1(0, 0, 1) : x_1, y_1, z_1 \in \mathbb{R}; x_1+y_1=0\}$

$F_2 = \{(x_2(1, 1, 0) + z_2(0, 1, 1) : x_2, y_2, z_2 \in \mathbb{R}; x_2+z_2=y_2\}$

So, the sum of 2 subspaces if I not mistakenly calculated is:

$F_1 + F_2 = \{(x_1+x_2, -x_1+x_2+z_2, z_1+z_2) \in \mathbb{R}^3, x_1+y_1=0;x_2+z_2=y_2\}$?

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  • $\begingroup$ That should be the sum, but perhaps there is an easier way to write it? $\endgroup$ – Michael Burr Feb 24 '17 at 12:31
  • $\begingroup$ Easier way to write it, what do you mean by that? $\endgroup$ – DomainFlag Feb 24 '17 at 12:33
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You should not mix parametric representation and comprehension, that is: describe a subspace of $\mathbb{R}^{3}$ either like a set of solutions of equations, either like a subspace generated by a family of vectors. For instance: $$\{(x,y,z):x+y=0\}=\{\alpha(1,-1,0)+\beta(0,0,1):\alpha \in \mathbb{R}, \beta \in \mathbb{R}\}.$$

Writing correctly things should help you.

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  • $\begingroup$ $$\{(x,y,z):x+y=0; \ x+z=y\}=\{\alpha(1,-1,0)+\beta(0,0,1)+\gamma(1,1,0)+\omega(0,1,1):\alpha \in \mathbb{R}, \beta \in \mathbb{R}, \gamma \in \mathbb{R}, \omega \in \mathbb{R}\}.$$ it should be like this written for F1 + F2??? $\endgroup$ – DomainFlag Feb 24 '17 at 12:46
  • $\begingroup$ Yes, that would be a good way to write $F_1+F_2$. $\endgroup$ – Michael Burr Feb 24 '17 at 12:57
  • $\begingroup$ okey thank you very much :) $\endgroup$ – DomainFlag Feb 24 '17 at 13:06
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Sketch:

  1. Observe first that $F_1$ contains the vectors $(1,-1,0)$ and $(0,0,1)$.

  2. Now, observe that $F_2$ contains the vectors $(1,1,0)$ and $(0,1,1)$.

  3. Therefore, $F_1+F_2$ contains all four of these vectors.

  4. One can show, with a little linear algebra (put the vectors in a matrix, row reduce the matrix and see if there's a pivot in every row), that these four vectors span all of $\mathbb{R}^3$. Therefore, $F_1+F_2=\mathbb{R}^3$.

Or, if you don't want to do row reduction, you can see that \begin{align*} \frac{1}{2}((1,-1,0)+(1,1,0))&=(1,0,0)\\ (0,1,1)-(0,0,1)&=(0,1,0)\\ (0,0,1)&=(0,0,1) \end{align*} so $F_1+F_2$ contains the standard basis for $\mathbb{R}^3$, so $F_1+F_2=\mathbb{R}^3$.

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