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:

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

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ $$\{(x,y,z):x+y=0; \ x+z=y\}=\{\alpha(1,-1,0)+\beta(0,0,1)+\gamma(1,1,0)+\omega(0,1,1):\alpha \in \mathbb{R}, \beta \in \mathbb{R}, \gamma \in \mathbb{R}, \omega \in \mathbb{R}\}.$$ 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


  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 $F_1+F_2=\mathbb{R}^3$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.