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Suppose [tex]W[/tex] and [tex]X[/tex] are subspaces of [tex]R^8[/tex].

Show that if dim[tex]W=3[/tex], dim[tex]X=5[/tex], and [tex]W+X = R^8[/tex], then [tex]W \cap X = \{0\}[/tex].

I can see this is obvious iff W and X are disjoint sets. If we add members of [tex]W[/tex] to [tex]X[/tex] in the usual way, and we get the new set [tex]W+X[/tex] whose dimension is now 8 (given), then the only way the dimension can be 8 is if each member of [tex]W[/tex] and [tex]X[/tex] is 'different', implying that [tex]W[/tex] and [tex]X[/tex] are disjoint and their intersection is zero.

Here is how I have tried to prove it...

1. We are given that [tex]R^8 = W + X[/tex]

2. I have assumed [tex]{W,X}[/tex] are independent - how would I prove this?

3. If (1) and (2) hold, then [tex]R^8[/tex] is also the direct sum of the two subspaces, [tex]R^8 = W \oplus X[/tex].

4. Let [tex]\textbf{v}[/tex] be a vector from [tex]W \cap X[/tex] and consider the equation [tex]\textbf{v} = \textbf{v}[/tex].

5. The left side of that equation is a member of [tex]W[/tex] and the right side is a linear combination of only one member of [tex]X[/tex]. But the independence of the spaces then implies that [tex]\textbf{v} = \textbf{0}[/tex].

6. Hence the intersection of [tex]W[/tex] and [tex]X[/tex] is trivial.

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# Intersection of disjoint SETS is empty

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