Direct sum complement is unique

  • Thread starter Bipolarity
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  • #1
Bipolarity
775
2
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
36,026
12,925
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
economicsnerd
269
24
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
Bipolarity
775
2
Thanks a lot guys! No wonder my proof has not been working out!
How might I prove that orthogonal complements are unique?

BiP
 
Last edited:
  • #5
economicsnerd
269
24
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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