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If A and B are vector spaces such that

A = B

Do they have the same dimension?

I think yes, because if they are equal then they can be spanned by some vectors. Any vectors that span A span B. Suppose then that A has a smaller dimension, meaning that it can be spanned by a smaller number of linearly independent vectors. Well, if these vectors span A, then they must be able to span B as well meaning that the dimensions are equal. Thus, A cannot have a smaller dimension (different dimension) than B.

Let me know if my logic is incorrect.

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Yes, your logic is correct. But much more generally, if two mathematical objects are equal, they must have the same properties. This follows from logic alone. –  Michael Greinecker Nov 12 '12 at 13:20
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If $A$ and $B$ are actually equal, then they're entirely the same, and there's no possible way that any property can be different from one to the other, so this kind of argument isn't necessary (although proving that the dimension is well-defined will look a bit like this). Did you mean that $A$ and $B$ are isomorphic? –  Matt Pressland Nov 12 '12 at 13:20
    
I just want to make sure that a n x n matrix A where nullspace(A) is equal to colspace(A) is not invertible. This is because if the null space and the column space were the same, they'd have the same dimension. Which means that the nullity(A) = dim[colspace(A)] = rank(A). By the Rank-Nullity Thm., rank(A) + nullity(A) = the number of columns (and in this case, rows). Which means that n = 2*rank(A) which means that there are some free variables which means that the matrix is not invertible. Does that make sense? –  M. Choi Nov 12 '12 at 13:27
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Yes, that makes. –  Berci Nov 12 '12 at 13:29
    
YES! Thank you. –  M. Choi Nov 12 '12 at 13:29

1 Answer 1

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Yes, correct, but the question seems too trivial.

Moreover, try to prove that there is a bijective linear mapping between vector spaces $A$ and $B$ if and only if $\dim A=\dim B$.

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The original exercise was not in the question, I just want to make sure that a n x n matrix A where nullspace(A) is equal to colspace(A) is not invertible. This is because if the null space and the column space were the same, they'd have the same dimension. Which means that the nullity(A) = dim[colspace(A)] = rank(A). By the Rank-Nullity Thm., rank(A) + nullity(A) = the number of columns (and in this case, rows). Which means that n = 2*rank(A) which means that there are some free variables which means that the matrix is not invertible. Does that make sense? –  M. Choi Nov 12 '12 at 13:28

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