Proving U1=U2 When U1, U2, W are Subspaces of V

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To prove that U1 equals U2 when both are subspaces of V with V expressed as a direct sum of W, it is essential to demonstrate that every vector in V has a unique representation. The proof begins by expressing any vector v in V as a sum of vectors from U1 and W, and similarly from U2 and W. By showing that the components from U1 and U2 must be equal due to the uniqueness of representation, it follows that U1 must equal U2. Explicitly stating that the vector from W remains consistent in both representations strengthens the argument. Using a basis for V to illustrate this uniqueness can further solidify the proof's validity.
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I have to prove or give a counter example to the statement if U1, U2, W are subspaces of V such that V=U1 direct sum W and V=U2 direct sum W, then U2=U1.

This is what I did: Let v be an element of V. Then v=v1+v2 for v1 an element of U1 and v2 and element of W and v=v3+v2 for v3 an element of U2. So v-v2=v1 and v-v2=v3. Therefore v1=v3. Hence U1=U2 since every vector in each subspace is the same.

I just feel like I am missing something to make my small proof 100% airtight. Should I mention somewhere that v is represented in a unique way since V=U1 direct sum W and V=U2 direct sum W?
 
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gravenewworld said:
This is what I did: Let v be an element of V. Then v=v1+v2 for v1 an element of U1 and v2 and element of W and v=v3+v2 for v3 an element of U2. So v-v2=v1 and v-v2=v3. Therefore v1=v3.
Be careful, how do you know whether your v2 in the expression v=v1+v2 is the same v2 as the one in v=v3+v2?
 
I guess I could explicity write that v2 is the same vector in both situations.
 
Actually, doesn't v2 have to be the same for both situations since v is the same? Since V is a direct sum of both subspaces then v has a unique representation so v2, has to be the same right?
 
Yes, that's true. But I would write that out explicitly, it doesn't appear trivial to me. I'd use a basis of V to write v, then take those vectors that are in U1 to form a basis for U1, the rest will form a basis for W. Then the result follows from the uniqueness of the basis expansion.
 
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