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Let $W_1,W_2,W_3$ be linear subspaces of finitely generated vector space V. Assume that $W_1\cap W_2=W_1\cap W_3=W_2 \cap W_3=\{\theta \}$. Is $W_1+W_2+W_3$ is always direct sum?

I think that the answer is NO and looking for a contradiction.

Please help, thank you!

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  • $\begingroup$ Is $\theta$ meant to be zero? $\endgroup$ Feb 18, 2014 at 20:01

2 Answers 2

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I'm assuming that by $\{ \theta \}$ , you meant $\{ \mathbf 0 \}$, the subspace consisting of just the zero vector.

In that case, the answer is yes, the sum is a direct sum (i.e., each thing in $W_1 + W_2 + W_3$ can be written in a unique way as $a + b + c$, where $a \in W_1$, $b \in W_2$, and $c \in W_3$).

Quick proof:

Lemma: If $U$ and $S$ are subspaces of $V$ with $U \cap S = \{ \mathbf 0 \}$, then $U + S$ is a direct sum

Proof: Suppose $x \in U + S$, and $x = u + s= u' + s'$, where $u, u' \in U$ and $s, s' \in S$. The $u + s = u' + s'$, so $u-u' = s - s'$. Since $u - u' \in U$, and $s - s' \in S$, we have that $u-u' = s-s' \in U \cap S = \{ \mathbf 0 \}$. Hence $u - u' = s - s' = 0$. Thus the decomposition of $x$ is unique.

Now: apply this lemma to $U = W_2$ and $S = W_3$ to conclude that every element of $Z = W_2 + W_3$ can be expressed in only one way as a sum of elements of $W_2$ and $W_3$.

Then observe that $W_1 \cap Z = W_1 \cap (W_2 + W_3) = (W_1 \cap W_2) + (W_1 \cap W_3) = 0$, and apply the lemma to $U = W_1$ and $S = Z$. Then any element in $U + Z = W_1 + W_2 + W_3$ can be expressed uniquely as $a + d$, where $a \in W_1$ and $d \in Z$. And $d$ can be expressed uniquely as $d = b+c$, where $b \in W_2$ and $c \in W_3$, by the prior application of the lemma.

We're done!

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Take $V=\mathbb R^2$ and $W_i = span\{ (1, i) \}$. Then the sum of the three subspaces is not a direct sum but they pairwise intersect only at the origin.

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