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The question that I am trying to solve is

Suppose that A is n × n and has rank n. What is its row space? What is its nullspace? What is its column space?

From rank nullity theorem, I know that the null space will be 0 because. Also, rank = dim(column space(A)) = dim(row space(A))

But the question asks for the row space and column space and not the dimensions of those. How can I find those two without knowing what the actual matrix is? Any help will be really appreciated. Thank you!

gazillionTickets
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    So you have a space $n$-vectors that has dimension $n$. What can that be? – Conifold Feb 18 '20 at 05:40
  • Umm an identity matrix? I dont know what you are trying to get at. From what i know , since its a non singular matrix, it can be reduced to an identity matrix with no free variables. – gazillionTickets Feb 18 '20 at 05:46
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    Forget about the matrix, all the useful information is already distilled into the question as I asked it. Suppose you have three linearly independent vectors in 3-space. What do they span? – Conifold Feb 18 '20 at 05:53
  • I think they span R^3. So by that logic the column and row space would be R^n? – gazillionTickets Feb 18 '20 at 05:58
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    Correct. Column space and null space are complementary (their sum is the whole space), the rank-nullity theorem tells you that the null space is $0$-dimensional, so it is just ${0}$. Its complement is the whole space. – Conifold Feb 18 '20 at 06:02
  • Thanks a lot! I still have one conceptual question. From the rank nullity theorem, we get to know that the dimension of row and column space is n. How does that imply that the 2 spaces span entire R^n. I know its probably a stupid question but I am trying to wrap my head around the difference between the dimension of a space and just the space itself. Thanks again – gazillionTickets Feb 18 '20 at 06:06
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    Because the dimension of $\mathbb{R}^n$ is also $n$. If you have $n$ vectors in it, and what they span has dimension $n$ they have no choice but to span it all. There are too many of them to span anything less. The dimension theorem says all bases have the same number of elements. Anything less than $\mathbb{R}^n$ must have lower dimension, so it can not contain $n$ independent vectors. – Conifold Feb 18 '20 at 06:13

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If $T: V\to V$, and the rank equals $n$, the dimension of $V$, then the range, column space, and row space are necessarily $V$.