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Just as a background information: I have absolutely no experience with topics in algebra. So please be patient, if I have done something terrible or not used the right terminology.

I have two vectorspaces containing vectors from a vectorspace as elements. The first vectorspace $A = \{v_2,v_3\}$ is a nullspace and the second vectorspace is $B = \{v_1,v_4\}$ and I need to evaluate the direct sum $A \oplus B=\{v_2,v_3\}\oplus\{v_1,v_4\}$.

This is what I have as a result (just by looking at other examples), but I am not sure if this is correct. It would be nice if someone more experienced could tell me if I did it right.

$$A \oplus B=\{v_2,v_3,v_2+\mu_1 v_3,v_1+\mu_1v_2+\mu_2v_3,v_4+\mu_1v_2+\mu_2v_3\}$$ where $\mu_1, \mu_2$ arbitrary numbers.

EDIT: The example that I have is from Ovsiannikov 1982 in which $\{v_4\}$ is the nullspace:

$$\{v_4\} \oplus \{v_1,v_2,v_1+v_3\}=\{v_1+\mu v_4,v_2+\mu v_4,v_1+v_3+\mu v_4,v_4\}$$

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    $\begingroup$ A direct sum is a direct sum of subvectorspaces. You have sets of vectors, not subspaces. $\endgroup$
    – Bernard
    Aug 3, 2016 at 19:33
  • $\begingroup$ Lets say $A$ and $B$ are subvectorspaces. I changed my question. So $A$ and $B$ are spanning a subvectorspace of the vectorspace. How do I procede? $\endgroup$
    – MrYouMath
    Aug 3, 2016 at 19:34
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    $\begingroup$ $A\oplus B$ is simply the set of all possible sums of a vector in $A$ and a vector in $B$. On checks easily these sums satisfy the definition of a subspace. $\endgroup$
    – Bernard
    Aug 3, 2016 at 19:37
  • $\begingroup$ So I need to add $v_2+\mu_1v_1+\mu_2v_4$, $v_3+\mu_1v_1+\mu_2v_4$ and so forth? Then shouldn't the result be just one vector $\mu_1v_1+\mu_2v_2+\mu_3v_3+\mu_4v_4$? $\endgroup$
    – MrYouMath
    Aug 3, 2016 at 19:40
  • $\begingroup$ The result is exactly what you say, if the vectors are bases of $A$ and $B$ respectively. $\endgroup$
    – Bernard
    Aug 3, 2016 at 19:44

1 Answer 1

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The direct sum is defined between two vector spaces $V,W$ over the same field $\mathbb{K}$. It is a new vector space whose elements are the couples $(v,w) : v \in V,w\in W$ withe the operations defined ''sede by side'': $(v,w)+(x,y)=(v+x,w+y)$ and $k(v,w)=(kv,kw)$.

So, in your case, we cannot define the direct sum between $A$ and $B$ but we can define the direct sum between $\overline{A}=\mbox{span}(A)$ and $\overline{B}=\mbox{span}(B)$ if $\mbox{span}(A) \cap \mbox{span}(A)=\{0\} $, i.e. if the elements of $A$ and $B$ are linearly independent.

In this case $\mbox{span}(A)\oplus \mbox{span}(B)$ is the vector space of the couples $(\mu_1 v_2+\mu_2 v_3,\lambda_1 v_1+\lambda_2v_4)$ and it is isomorphic to the space spanned by $(v_1,v_2,v_3,v_4)$ if these are vectors of the same vector space.

Your ''result'' seems to be an intuition of this fact, but with some confusion.

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  • $\begingroup$ 1+ for your efforts, But I think that is not what I am looking for. I added the example that I used to understand what is done. $\endgroup$
    – MrYouMath
    Aug 3, 2016 at 19:54

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