Suppose we have a matrix $A$ with $m$ rows and $n$ columns satisfying the condition $m<n$. Suppose further that $m$ rows are linearly independent, and $n$ columns are linearly independent as well. The article from Wikipedia says:

In linear algebra, the rank of a matrix $A$ is the dimension of the vector space generated (or spanned) by its columns. This corresponds to the maximal number of linearly independent columns of $A$. This, in turn, is identical to the dimension of the space spanned by its rows.

Definitely, from the exmaple above we have $rank(A)=m$. But from another hand, the dimension of the space spanned by rows are equal to $n$, since $n$ linearly independent equations. If my reasoning is correct, I see a contradiction between the following property $rank(A)=min(m,n).$ Please guide me what part of my reasoning is incorrect.

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    $\begingroup$ Did you mean to type that $n$ columns are linearly independent? (There are only $m<n$ rows.) This is the mistake. Precisely $m$ of the column vectors will be linearly independent. $\endgroup$ – Ted Shifrin Sep 13 '19 at 16:34
  • $\begingroup$ @TedShifrin Corrected. Thanks! $\endgroup$ – sane Sep 13 '19 at 16:35
  • $\begingroup$ Suppose $m=1$ and $n=2$, i.e. $A$ is a row vector with two entries. Can you give an example of such an $A$ whose two columns are linearly independent? $\endgroup$ – user1551 Sep 13 '19 at 16:39
  • $\begingroup$ You said m<n and m rows are linearly independent (this would also mean m, and not n, columns are linearly independent). $\endgroup$ – BJKShah Sep 13 '19 at 16:53

You assumed that it was possible for all the $n$ columns to be linearly independent. If each column only has $m$ entries and $m<n$, then that's impossible (you can't, for instance, have three linearly independent $2\times 1$ columns).

In fact, it turns out that the largest size of a set of linearly independent columns you can get from a given matrix is equal to the largest size of aset of linearly independent rows you can get. This number is what is called the rank of that matrix.

  • $\begingroup$ Thank you for your answer. If $rank(A)=m$ this means that we have $m$ linearly independent columns. Does this presume, that we have $m$ linearly independent rows as well? This part is confusing for me. Thank you! $\endgroup$ – sane Sep 13 '19 at 16:44
  • $\begingroup$ @sane "$\operatorname{Rank}(A) = k$" is equivalent to "there are $k$ linearly independent rows in $A$, but not $k+1$" is equivalent to "there are $k$ linearly independent columns in $A$, but not $k+1$", no matter what $k$ is in relation to $m$ and $n$. Any of the three statements implies the other two. If $k$ happens to be $m$, then yes, that too. $\endgroup$ – Arthur Sep 13 '19 at 16:48
  • $\begingroup$ From your given definitions follows: "there are $m$ independent rows (from my assumption)", and "there are $n$ independent columns" (from my assumption). Is this correct? I don't understand also the following "you can't, for instance, have three linearly independent 2\times1 columns", why? Thank you, sorry for many questions. $\endgroup$ – sane Sep 13 '19 at 17:03
  • $\begingroup$ @sane If $A$ is $2\times 3$, then you can have two independent rows. In that case, you also have two independent columns, and we say that the rank is $2$. However, you absolutely cannot have three independent columns. That's impossible. $\endgroup$ – Arthur Sep 13 '19 at 17:56
  • $\begingroup$ Again thank you very much for your comment. Suppose we have $x_1, x_2, x_3$ vectors, each of them are $2 \times 1$. From the definition of linear dependence, we have: if exists $\alpha_1, \alpha_2, \alpha_3$ scalars, s.t. holds $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=0$. Obviously, the later makes a underdetermined system of linear equations, therefore we can have either zero or infinitely many solutions. This system of linear equations should have zero solution in order to have linearly independent vectors. But infinitely many solutions are feasible as well, implying lin dependence. Am I wrong? $\endgroup$ – sane Sep 13 '19 at 19:35

if $m<n$. and if the m rows are linearly independent, then n columns cannot be linearly independent.(row rank=column rank, for any matrix) So no contradiction. The definition is correct.

  • $\begingroup$ Why we can't have simultaneously $m$ linearly independent rows and $n$ linearly independent columns in matrix, satisfying $m<n$? $\endgroup$ – sane Sep 13 '19 at 17:36

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