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If $W_1,\dots, W_k$ are subspaces of a finite dimensional vector space $V$ such that $W_1+\cdots+W_k=V$, and I want to show that $V=W_1\oplus\cdots\oplus W_k$ if and only if $\dim(V)=\sum{W_i}$, then will what's displayed below suffice?


$$V=W_1\oplus\cdots\oplus W_k$$ $$\iff$$ $$V=W_1+\cdots+W_k~\text{and}~W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$$ $$\iff$$ $$\text{The subspaces $W_i$ are independent; that is, no sum $w_1+\cdots+w_k$ with $w_i$ in $W_i$ is zero except the trivial sum.}$$ $$\iff$$ $${\scr{B}}=\{\beta_1,\dots,\beta_k\}~\text{is a basis for $V$, where $\beta_i$ is a basis for $W_i$}$$ $$\iff$$ $$\dim{V}=\dim{(W_1+\cdots+W_k)}=\dim{W_1}+\cdots+\dim{W_k}=~\mid\beta_1\mid+\cdots+\mid\beta_k\mid=k$$ $$\iff$$ $$\overset{\text{Does this belong here?}}{\dim{\scr{B}}=k}$$

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  • $\begingroup$ you are supposing that $W_i$ $i\in {1,2,..,k}$ has dimension $1$ wich must not be true $\endgroup$ – sigmatau May 22 '13 at 20:50
  • $\begingroup$ Which denotes cardinality then? $\endgroup$ – Trancot May 22 '13 at 20:52
  • $\begingroup$ Yes, that is correct. $|A|$ denotes the cardinality of set $A$. If $\beta_i$ is a basis for $W_i$, then $|\beta_i|=\dim W_i$. $\endgroup$ – Jared May 22 '13 at 20:53
  • $\begingroup$ The second line shows a big misunderstanding of the direct sum... it is not equivalent to being a direct sum. For instance, suppose $A\cap B=\{0\}$. Then $A\cap A\cap B=\{0\}$ also but the sum of $A+A+B$ is not direct. The correct version is: $(\sum_{i\neq j} W_i)\cap W_j=\{0\}$. $\endgroup$ – rschwieb May 22 '13 at 21:08
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    $\begingroup$ Try to reduce your usage of notation. In particular avoid using $\iff$ between statements that are not strictly equivalent. None of the equivalences in your post actually is an equivalence of mathematical statements. $\endgroup$ – Martin May 22 '13 at 21:12
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Your first equivalence is false: when $k \geq 3$, $\bigcap_{i=1}^k W_k = \{0\}$ is too weak to imply that the spaces $W_1,\ldots,w_k$ are independent, i.e., that $W_1 + \ldots + W_k = W_1 \oplus \ldots \oplus W_k$. For example, take $V = \mathbb{R}^2$, $W_1 = \langle (1,0) \rangle$, $W_2 = \langle (0,1) \rangle$, $W_3 = \langle (1,1) \rangle$. This is one of two or three subtle traps in linear algebra that even professional mathematicians can fall into if they're not careful.

Any of the following is an acceptable definition of independent subspaces (and they are equivalent):

(i) For all $1 \leq i \leq k$, $W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$.
(ii) If for all $i$ we choose a nonzero $v_i \in W_i$, then $\{v_1,\ldots,v_k\}$ is a linearly independent set.
(iii) If for all $i$ we choose a linearly independent set $S_i \subset W_i$, then $S = \bigcup_{i=1}^k S_i$ is a linearly independent set.

Note that if you believe that (iii) is equivalent to the sum being a direct sum then you've got the equivalence you're asking about, so I suggest you concentrate instead on showing these conditions are equivalent to your given definition of internal direct sums.

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  • $\begingroup$ So what about this then? $\endgroup$ – Trancot May 22 '13 at 21:45
  • $\begingroup$ @Barisa: your new version is improved. If you take your first $\iff$ as the definition of the sum being direct, it seems to me that you should explain why the third $\iff$ holds (it's hard for me to evaluate the second $\iff$ because I don't know which definition of "independent subspaces" you're taking). If that $\iff$ can be taken as known, then your argument is essentially complete. $\endgroup$ – Pete L. Clark May 22 '13 at 23:14
  • $\begingroup$ What about now? $\endgroup$ – Trancot May 23 '13 at 17:28

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