My book writes:

Definition. A vector space $V$ is called the direct sum of $W_1$ and $W_2$ if $W_1$ and $W_2$ are subspaces of $V$ such that $W_1 \cap W_2=\{0\}$ and $W_1 + W_2 = V$. We denote that $V$ is the direct sum of $W_1$ and $W_2$ by writing $V=W_1\oplus W_2$.

I'm not sure what I should imagine $W_1 + W_2 = V$ as.

Thank you for any help !

  • 1
    $\begingroup$ It should be $W_1\cap W_2$ instead of $W_1\cup W_2$. $\endgroup$
    – joriki
    Sep 27 '12 at 10:46
  • $\begingroup$ A free book by Jim Hefferon on Linear Algebra at joshua.smcvt.edu/linearalgebra Page 129 has a good explanation $\endgroup$
    – Vikram
    Sep 27 '12 at 11:13

Take this example to clarify the difference: $$V=\mathbb{R}^{2}$$ $$W_{1}=sp_{\mathbb{R}}\{(1,0)\}=\{(a,0)|a\in\mathbb{R}\}$$ $$W_{2}=sp_{\mathbb{R}}\{(0,1)\}=\{(0,b)|b\in\mathbb{R}\}$$


$$W_{1}+W_{2}=\{w_{1}+w_{2}|w_{i}\in W_{i}\}=\{(a,0)+(0,b)|a,b\in\mathbb{R}\}=\{(a,b)|a,b\in\mathbb{R}\}$$


$$W_{1}\cup W_{2}=\{v_{1}|v_{1}\in W_{1}\}\cup\{v_{2}|v_{2}\in W_{2}\}$$ and this set is consistent of all elements of the form $(a,0)$ and $(0,b)$ (where $a,b\in\mathbb{R})$ but, for example, $(1,1)\not\in W_{1}\cup W_{2}$.

  • $\begingroup$ thank you ! would you be so kind to tell me what you mean by $sp_{\mathbb{E}}$ though? :P $\endgroup$
    – rollover
    Sep 27 '12 at 10:58
  • $\begingroup$ @foaly - sorry, it was a typo. is it clear now ? $\endgroup$
    – Belgi
    Sep 27 '12 at 10:59
  • $\begingroup$ no.. I don't know what that $sp$ think means :P probably some kind of notation not known to me (yet) ? $\endgroup$
    – rollover
    Sep 27 '12 at 11:02
  • $\begingroup$ it means span. do you know what this means ? $\endgroup$
    – Belgi
    Sep 27 '12 at 11:05
  • $\begingroup$ oh ya. the notations i know are $\\span(1,0)$ and $<1,0>$ (or similar to that). Why do you write the index $\mathbb{R}$ to it? $\endgroup$
    – rollover
    Sep 27 '12 at 11:06

$$W_1+W_2=\{w_1+w_2\mid w_1\in W_1\land w_2\in W_2\}\;.$$

  • $\begingroup$ Elegant and simple. $\endgroup$ Jan 7 '18 at 15:06

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.