- #1
jecharla
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Exercise #17 in Linear Algebra done right is to prove that the dimension of the direct sum of subspaces of V is equal to the sum of the dimensions of the individual subspaces. I have been trying to figure this out for a few days now and I'm really stuck. Here's what I have got so far:
Choose a base of each of the subspaces and combine them in one list which clearly spans V and has the length we are looking for. Now we just have to show that this list is linearly independent.
If this list is not linearly independent than one of the vectors is in the span of the previous vectors. If I could show that this vector must be in the span of one of the other bases, then that would prove the result but I can't quite figure out how to get there. Any help would be greatly appreciated.
Choose a base of each of the subspaces and combine them in one list which clearly spans V and has the length we are looking for. Now we just have to show that this list is linearly independent.
If this list is not linearly independent than one of the vectors is in the span of the previous vectors. If I could show that this vector must be in the span of one of the other bases, then that would prove the result but I can't quite figure out how to get there. Any help would be greatly appreciated.