Direct sum complement is unique

  • Thread starter Bipolarity
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  • #1
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Main Question or Discussion Point

I'm curious about whether a statement I conjecture about direct sums is true.
Suppose that ##V## is a finite-dimensional vector space and ##W##,##W_{1}##,##W_{2}## are subspaces of ##V##. Let ## V = W_{1} \bigoplus W ## and ## V = W_{2} \bigoplus W ##.

Then is it the case that ## W_{1} = W_{2} ##?

I merely need to know whether this is true or not so that I can know which direction to steer my proof. I am guessing it is true, but am having trouble proving it, and that is giving me doubts as to whether or not it is true.

All help is appreciated! Thanks!

BiP
 

Answers and Replies

  • #2
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I am not sure how your direct sum is evaluated. What about this?
V=R2
W={(x,0)}, W1={(x,x)}, W2={(x,-x)}
 
  • #3
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In the category of vector spaces: no, as mfb showed.
In the category of inner-product spaces: yes. In this case, we say [itex]V=W_i\oplus W[/itex] if [itex]V=W_i+W[/itex] and [itex]W_i, W[/itex] are orthogonal.
 
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  • #4
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Thanks a lot guys! No wonder my proof has not been working out!
How might I prove that orthogonal complements are unique?

BiP
 
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  • #5
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Suppose [itex]V=W_1+ W[/itex] and [itex]W_1 \perp W[/itex]. Let [itex]W_2 = \{v\in V: \enspace v\perp W\}[/itex].

By construction, [itex]W_2 \supseteq W_1[/itex]. Try to show that [itex]W_1, W_2[/itex] have the same (finite) dimension... hint: dimension theorem. Then use that no finite-dimensional vector space has a proper subspace of the same dimension.
 

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