Rank of a Matrix - Demonstration or Postulate? - I realized that the teacher  - I do not know why, but this one - obviously - is not the question - takes demonstrations he shows during lessons and I usually do not succeed in understanding from an old Linear Algebra's book. 
I hope that somebodby will be able to  help me on an important theorem still leaving me with more doubts than certainties. The theorem text is as follows: 
"A matrix has rank k if - and only if - it has k rows - and k columns - linearly independent, whilst each one of the remaining rows - and columns - is a linear combination of the k preceding ones".
Let's suppose that the matrix is $$\begin{bmatrix}a_{1,1}&a_{1,2}& \ldots&a_{1,n}\\a_{2,1}&a_{2,2}&\ldots&a_{2,n}\\\ldots&\ldots&\ldots&\ldots\\a_{m,1}&a_{m,2}&\ldots&a_{m,n}\end{bmatrix}\tag{1}$$ and let's suppose that the minor $$\lambda = \begin{vmatrix}a_{1,1}&a_{1,2}&\ldots&a_{1,k}\\a_{2,1}&a_{2,2}&\ldots&a_{2,k}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&a_{k,2}&\ldots&a_{k,n}\end{vmatrix}\tag{23}$$ formed by the first $k$ rows and the first $k$ columns - situation which it is always possible to get to by means of adequate substitutions between rows and columns - is not null. It should be noted  that the first $k$ rows are linearly independent. Indeed, should the contrary be true, at least one of the rows would be a linear combination of the remaining ones and minor $(23)$ would be null against the hypothesis. Let's now consider minor $(24)$ - having order $k+1$ - $$\begin{vmatrix}a_{1,1}&\ldots&a_{1,k}&a_{1,j}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}&a_{k,j}\\a_{i,1}&\ldots&a_{i,k}&a_{i,j}\end{vmatrix}\tag{24}$$ with $i = k+1,k+2, \ldots, m$ and $j = 1,2, . . ., n$. This minor is null: if $j=<k$ because it happens to feature two identical columns, but even if $j>k$. Actually, determinant $(24)$ can be developed according to the elements of the last column and one gets: $$\lambda a_{i,j}+\lambda_1a _{1,j}+\lambda_2a_{2,j}+ \ldots +\lambda_ka_{k,j}=0\tag{25}$$ where $$\lambda_1,\lambda_2,\ldots, \lambda_k\tag{26}$$ are nothing else but the algebraic complements - with respect to determinant $(24)$ - of $a_{1,j}, a_{2,j}, \ldots, a_{k,j}$. It should be explicitly noted that $(26)$ are constants and that they do not depend on $j$ - WHY? - . After solving $(25)$ with respect to $a_{i,j}$ - remember that $\lambda$ is not null - and putting $\mu_1= -\lambda_1/\lambda$, $\mu_2= -\lambda_2/\lambda, \ldots, \mu_k= -\lambda_k/\lambda$, one gets $a_{i,j}=\mu_1a _{1,j}+\mu_2a_{2,j}+ \ldots +\mu_ka_{k,j}$ with $1\le j\le n$. As a consequence, the  $i$-th row ($i>k$) in the matrix is nothing else but a linear combination of the first $k$ ones . The theorem is thus proved as for raws and can be analogously demonstrated as for columns. 
Apparently everything seems so simple and trivial. But I cannot understand all the same. I think that assuming coefficients $\lambda_1, \lambda_2, \ldots, \lambda_k$ - algebraic complements with respect to the determinant $(24)$ - do not depend on $j$ -  that is on the column, but only on $i$, that is on the row beyond the rank, even if the text of the demonstration does not explicitly admits it - is totally equivalent to the demonstration itself. But - even more - I do not understand at all why the algebraic complements of the same index should always be equal in the corresponding positions independently on the columns on which one calculates the determinant. Based on what? On which theorem ? All the theorem is based on the fact that coefficients $\lambda$ depend only on the raw to which they belong and that , as a consequence, if they belong to the same raw, even if the column varies, they are equal. In fact, if they - generally - depended on $\{i,j\}$  - that is if they were $\lambda_{i,j}$ as I would have expected them to be and not simply $\lambda_i$ -, the theorem would not be valid any longer. What's there so evident and, still, thoroughly escaping me? 
Thank you so much!
 A: Simply write down carefully the development of the determinant $(24)$:
$$\begin{vmatrix}a_{1,1}&\ldots&a_{1,k}&a_{1,j}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}&a_{k,j}\\a_{i,1}&\ldots&a_{i,k}&a_{i,j}\end{vmatrix} = \lambda a_{i,j}+\begin{vmatrix}a_{2,1}&\ldots&a_{2,k}\\ \ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}\\a_{i,1}&\ldots&a_{i,k}\end{vmatrix}a_{1,j}+ \dots\tag{24-25}$$
You see that $\lambda_1 = \begin{vmatrix}a_{2,1}&\ldots&a_{2,k}\\ \ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}\\a_{i,1}&\ldots&a_{i,k}\end{vmatrix}$ does not depend on $j$ (it's made with the lines $2$ to $k$ and $i$ and the columns $1$ to $k$) and it's the same for the others.
A: I realized that the teacher - I do not know why, but this one - obviously - is not the question - takes demonstrations he shows during lessons and I usually do not succeed in understanding from an old Linear Algebra's book.
I hope that somebodby will be able to help me on an important theorem still leaving me with more doubts than certainties. The theorem text is as follows:
"A matrix has rank k if - and only if - it has k rows - and k columns - linearly independent, whilst each one of the remaining rows - and columns - is a linear combination of the k preceding ones".
Let's suppose that the matrix is
$$\begin{bmatrix}a_{1,1}&a_{1,2}& \ldots&a_{1,n}\\a_{2,1}&a_{2,2}&\ldots&a_{2,n}\\\ldots&\ldots&\ldots&\ldots\\a_{m,1}&a_{m,2}&\ldots&a_{m,n}\end{bmatrix}\tag{1}$$
and let's suppose that the minor
$$\lambda = \begin{vmatrix}a_{1,1}&a_{1,2}&\ldots&a_{1,k}\\a_{2,1}&a_{2,2}&\ldots&a_{2,k}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&a_{k,2}&\ldots&a_{k,n}\end{vmatrix}\tag{23}$$
formed by the first k rows and the first k columns - situation which it is always possible to get to by means of adequate substitutions between rows and columns - is not null. It should be noted that the first k rows are linearly independent. Indeed, should the contrary be true, at least one of the rows would be a linear combination of the remaining ones and minor (23) would be null against the hypothesis. Let's now consider minor (24) - having order k+1 - $$\begin{vmatrix}a_{1,1}&\ldots&a_{1,k}&a_{1,j}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}&a_{k,j}\\a_{i,1}&\ldots&a_{i,k}&a_{i,j}\end{vmatrix}\tag{24}$$
with $i=k+1,k+2,…,m$ and $j=1,2,...,n$. This minor is null: if $j=<k$ because it happens to feature two identical columns, but even if $j>k$. Actually, determinant (24) can be developed according to the elements of the last column and one gets:
$\lambda a_{i,j}+\lambda_1 a_{1,j}+\lambda_2 a_{2,j}+ . . . + \lambda _k a_{k,j} = 0 $ (25)
where}
$\lambda_1, \lambda_2, . . ., \lambda_k$ (26)
are nothing else but the algebraic complements - with respect to determinant (24) - of $a_{1,j},a_{2,j}, . . .,a_{k,j}$. It should be explicitly noted that (26) are constants and that they do not depend on $j$ - WHY? - . After solving (25) with respect to $a_{i,j}$ - remember that $\lambda$ is not null - and putting $\mu_1 = − \lambda_1/\lambda, \mu_2 = − \lambda_2/\lambda, . . ., \mu_k = − \lambda_k/\lambda$, one gets $a_{i,j} = \mu_1 a_{1,j} + \mu_2 a_{2,j} + . . . + \mu_k a_{k,j}$ with $1≤j≤n$. As a consequence, the $i$-th row ($i>k$) in the matrix is nothing else but a linear combination of the first $k$ ones.
The theorem is thus proved as for raws and can be analogously demonstrated as for columns.
Apparently everything seems so simple and trivial. But I cannot understand all the same. I think that assuming coefficients $\lambda_1, \lambda_2, . . ., \lambda_k$ - algebraic complements with respect to the determinant (24) - do not depend on $j$ - that is on the column, but only on $i$, that is on the row beyond the rank, even if the text of the demonstration does not explicitly admits it - is totally equivalent to the demonstration itself. But - even more - I do not understand at all why the algebraic complements of the same index should always be equal in the corresponding positions independently on the columns on which one calculates the determinant. All the theorem is based on the fact that coefficients $\lambda$ depend only on the raw to which they belong and that , as a consequence, if they belong to the same raw, even if the column varies, they are equal. In fact, if they - generally - depended on $(i,j)$ - that is if they were $\lambda_{i,j}$ as I would have expected them to be and not simply $\lambda_i$ -, the theorem would not be valid any longer.
If one develops the determinant (24 )- as suggested by paf -: $$\begin{vmatrix}a_{1,1}&\ldots&a_{1,k}&a_{1,j}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}&a_{k,j}\\a_{i,1}&\ldots&a_{i,k}&a_{i,j}\end{vmatrix} = \lambda a_{i,j} + \begin{vmatrix}a_{2,1}&\ldots&a_{2,k}\\a_{k,1}&\dots&a_{k,k}\\a_{i,1}&\ldots&a_{i,k}\end{vmatrix} a_{1,j} + . . . \tag{24-25} $$, one can realize that
$$\lambda_1 = \begin{vmatrix}a_{2,1}&\ldots&a_{2,k}\\a_{k,1}&\dots&a_{k,k}\\a_{i,1}&\ldots&a_{i,k}\end{vmatrix}$$ does not depend on $j$ (it's, obviously, made with the lines $2$ to $k$ and $i$ and the columns $1$ to $k$).
But - formally, at least - it's not the same for the other columns. Hence the reason for repeating my question. If one calculates $\lambda$'s on other columns, it proves impossible to manage to formally show that - for the same index - they are equal and this - nevertheless - is what a demonstration should grant. This is the reason why I think that the answer I got is correct and useful, but did not take into consideration the totalness of my doubt. I still wonder whether this one can be correctly deemed of as an effective demonstration, as its author wants to propose it, or whether it should simply be retained as a sort of skilful verification, but only a verification and not a real demonstration.
In fact, it is not sufficient to only prove that $\lambda$'s are not dependent on $j$'s, but it should be proved that they are equal when calculated on different columns of the matrix.
