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Suppose that A is an $m$ by $n$ matrix of real numbers. How can I show that the dimension of the span of the columns of A (in $\mathbb{R}^m$) equals the dimension of the span of the rows of A (in $\mathbb{R}^n$)?

This is a question from my textbook and it relies heavily on other exercises such that in the end, to prove this one question you must prove at least 3 other exercises. I was wondering if a self-contained shorter proof was available. Thanks!

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Try looking at the reduced row echelon form... –  Tarnation Mar 19 at 2:17
    
which text book? –  Brad S. Mar 19 at 4:33
    
Linear algebra done right by Sheldon axler –  user123276 Mar 19 at 5:37

1 Answer 1

up vote 2 down vote accepted

This result relies on (several) other results regardless of how you approach the proof, so if you have to prove it (AND all the theorems it relies on) then it will be some work. Anyway - here's two approaches I can think of:

1> This will work for matrices over a general field: you need the following

  • For nonsingular $P$ and $Q$: dim(CS$(PAQ))=$dim(CS$(A))$
  • For nonsingular $P$ and $Q$: dim(RS$(PAQ))=$dim(RS$(A))$
  • For any $A$ there exists nonsingular $P$ and $Q$ so that $PAQ=\begin{bmatrix}I_r & 0 \\ 0 &0 \end{bmatrix}$, with $r$ unique

Unfortunately all of the above itself requires some work to prove. But once you have these three it is easy to prove the result you require.

2> Now another approach: using the standard inner product of $\mathbb{R}^k$, you just need the following:

  • The nullspace of $A$ is the orthogonal complement of the rowspace of $A$ (this is easy to prove).
  • The dimension of the nullspace of $A$ plus the dimension of the column space of $A$ equals the number of columns of $A$ (which is the dimension of $\mathbb{R}^n$) (this is harder to prove but do-able)

Combine the two above and you have the required result

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