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Linear algebra direct sum proof

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zcd
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Homework Statement


Let W1 and W2 be subspaces of a vector space V. Prove that [tex]W_1\oplus{}W_2=V \iff[/tex] each vector in V can be uniquely written as x1+x2=v, where [tex]x_1\in W_1[/tex] and [tex]x_2\in W_2[/tex]


Homework Equations


[tex]W_1\oplus{}W_2=V[/tex] means [tex]W_1\cap W_2 =\{0\}[/tex], [tex]W_1 + W_2 =V[/tex] and W1 & W2 are subspaces of V

8 axioms defining vector space


The Attempt at a Solution


I'm trying to assume that [tex]\exists x'_1,x'_2: x'_1+x'_2=v[/tex] and [tex]x'_1\in W_1, x'_2\in W_2[/tex] and then proving [tex]x'_1=x_1, x'_2=x_2[/tex], but I'm unsure of where to go from that step.
 
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Answers and Replies

  • #2
Dick
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If x1+x2=v and x1'+x2'=v then (x1-x1')+(x2-x2')=0. (x1-x1') is in W1. (x2-x2') is in W2. 0 is in W1. Is (x2-x2') in W1?
 
  • #3
zcd
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[tex](x_2-x'_2)\in W_1\iff x_2-x'_2=0[/tex]
 
  • #4
Dick
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[tex](x_2-x'_2)\in W_1\iff x_2-x'_2=0[/tex]
So you've got it?
 
  • #5
zcd
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I can see how this proves [tex]W_1\oplus{}W_2=V[/tex], but I'm still unsure on how x1 and x2 are unique. Is it because if [tex]x_i\neq x'_i[/tex] then the two vectors are in different subspaces, while if [tex]x_i= x'_i[/tex] the two subspaces will overlap at zero?
 
  • #6
Dick
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I can see how this proves [tex]W_1\oplus{}W_2=V[/tex], but I'm still unsure on how x1 and x2 are unique. Is it because if [tex]x_i\neq x'_i[/tex] then the two vectors are in different subspaces, while if [tex]x_i= x'_i[/tex] the two subspaces will overlap at zero?
It doesn't prove W1+W2=V. It proves if W1+W2=V then the decomposition is unique. You have to also prove if the decomposition is unique then W1+W2=V.
 
  • #7
zcd
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For the reverse argument, can I say that [tex]S:=W_1 \cap W_2[/tex] and [tex]\exists s\inS: x_1+x_2+s=v\implies s=0[/tex], therefore [tex]S=\{ 0\}[/tex]
 

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