# Direct sums of subspaces

Can someone check the correctness of my proof.

Statement.

1. A single subspace $W_1$ is independent.
2. Two subspaces $W_1,W_2$ are independent $\iff$ $W_1\cap W_2=\{0\}$

Two subspaces are said to be independent if $w_1+w_2=0, w_1\in W_1, w_2\in W_2$ implies that $w_1=0$ and $w_2=0$.

Proof.

1. $0$ can be expressed uniquely once a basis for $W_1$ is chosen.
2. Let $w\in W_1\cap W_2$ and let $B=(v_1,\dots,v_n),C=(w_1,\dots,w_m)$ be bases for $W_1,W_2$. Then $$w=\sum_i x_i v_i=\sum_j y_j w_j$$ $$0=w-w=\sum_i x_i v_i-\sum_j y_j w_j$$ As $B,C$ are bases, we have: $$x_i=0\text{ }\forall i, y_j = 0\text{ } \forall j$$ Substituting back we get $w=0$ hence $W_1\cap W_2=\{0\}$

Conversely, suppose $W_1\cap W_2=\{0\}$. Consider the linear relation $$\sum_i x_i v_i + \sum_j y_j w_j$$ $$\sum_i x_i v_i = \sum_j (-y_j) w_j=v, \text{ for some } v$$ So $v=0\implies x_i = 0$ and $y_j=0$ for all $i,j$ because $B,C$ are bases. Therefor, $W_1,W_2$ are independent subspaces.

This argument extends to arbitrary collections of subspaces (by induction).

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As far as I know, there is no sense in defining independence of a single subspace. If you are doing it for the sake of induction, start your induction at $2$. – Vishal Gupta Aug 27 '13 at 13:46
For the second case your argument seems to be fine. But what is your induction statement? – Vishal Gupta Aug 27 '13 at 13:47
@Vishal The intersection of n subspaces is 0 iff they are independent – saadtaame Aug 27 '13 at 13:53
The argument does not extend to more than 2 subspaces. For example, 3 lines in a 2-dimensional space are never independent. – Etienne Aug 27 '13 at 14:23
@Etienne Okay. Thank you! – saadtaame Aug 27 '13 at 18:53

Your proof for (2) is not entirely correct (but you're on the right path!). From $$0 = \sum_ix_iv_i - \sum_jy_jw_j,$$ you cannot conclude that $x_i=y_j=0$ for all $i,j$ from the fact that $\{v_i\}$and $\{w_j\}$ are bases of $W_1$ and $W_2$ respectively. This assumes that $(v_1,\dots,v_n,w_1,\dots,w_n)$ is itself a basis... which you cannot assume at this point. Indeed, you're trying to show that $W_1\cap W_2=\{0\}$.

Notice that $\sum_ix_iv_i\in W_1$ and $\sum_j(-y_j)w_j\in W_2$. As $W_1$ and $W_2$ are independent, this implies that $$\sum_ix_iv_i = \sum_j(-y_j)w_j = 0$$ and so $w=0$. Therefore, $W_1\cap W_2=\{0\}$.
We consider the linear relation $v+w=0$ for some $v\in W_1$ and $w\in W_2$. Then $v=-w$. We can conclude that $v,w\in W_1\cap W_2$. By hypothesis $v=w=0$. By definition, $W_1$ and $W_2$ are independent.
Note that you began by writing a sum (expression) in $W_1+W_2$, but not a linear relation. Then how did you find that $\sum_ix_iv_i=\sum_j(-y_j)w_j$? And when you set $v=0$, you were essentially assuming that the summands were both $0$.
To see why your final remark on the generalization is false, you might try to find subspaces $W_1,W_2,W_3$ such that $\bigcap_iW_i=\{0\}$ but not all of $$W_1\cap W_2,\, W_1\cap W_3,\text{ and } W_2\cap W_3$$ are trivial intersections.