I was wondering if this reasoning could be considered a proof:

Let $A$ be a matrix $(m \times n)$ with $m$ rows and $n$ columns. Now we consider that there are at least $k$ rows vectors that are linearly independent, and that the span of $(v_1,v_2,...,v_k)$ row vectors includes the space of columns. So the dimension of columns is less or equal of the dimension of rows $\text{span}(v_1,v_2,...,v_k)$.

By applying the same reasononing to columns, let the columns have $r$ linearly independent vectors $(v_1,v_2,...,v_r)$ the span of columns $\supseteq$ space of rows, and therefore the dimension of rows is less or equal to the dimension of span of columns vectors $(v_1,v_2,...,v_r)$ therefore to respect both of equations dimension of column is equal of the dimension of rows.

My reasoning is probably wrong but I would like to get an advice :).


Here's a $1 \times 3$ matrix: $$ A = \pmatrix{1 & 3 & 2} $$ of row-rank $1$. Your argument says that if I look at one linearly independent row (that'd be row 1, of course), it "includes the space of columns".

Now the column-space of this matrix is just $\Bbb R$, so that claim doesn't seem to make sense.

So I'd say that your proof is flawed.

Broad hint: It's often wise to sanity-check things on low-dimensional examples.

| cite | improve this answer | |
  • $\begingroup$ Sorry I problably didn't explain myself that well(english is not my main language),what I meant is that you should look at all the linearly independent rows example: kv1+kv2=v3 where v1,v2,v3 are vectors of 3 elements (1x,2y,3z),(1x,3y,2z),(2x,5y,5z) then by looking just at first ,second and third columns we can see that columns vectors got 'generated'.In the example you proposed by looking at the vector columns we can see that the dim(1),(3),(2) is 1 cuz 1*1+1*3-2*2=0 so they are linearly dependent,by using the fact that span of columns ⊇ space of rows and span of rows ⊇ space of columns $\endgroup$ – Argenti21 Jul 11 at 21:01
  • $\begingroup$ we can say that dimension=1 $\endgroup$ – Argenti21 Jul 11 at 21:01
  • $\begingroup$ Yeah...but if there were two rows, how do you know that the columns corresponding to these rows would have two independent elements rather than just one? Answer: you prove this theorem! What you've got so far is really just handwaving. $\endgroup$ – John Hughes Jul 11 at 21:18
  • $\begingroup$ ahah I get your point,it's the first time I tried to do something like that , Thank you for helping @Jogn Hughes $\endgroup$ – Argenti21 Jul 11 at 21:29
  • $\begingroup$ Happy to help. It's one of those theorems that feels as if it should be easy, and it's not too hard, but one's always looking for a simpler proof... $\endgroup$ – John Hughes Jul 12 at 0:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.