Understanding Rank of a Matrix: Important Theorem and Demonstration

[Conrad Manfried]

It is the demonstration of an important theorem I do not succeed in understanding.

"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:
λa_i,j+λ₁a_1,j+λ₂a_2,j+ . . . + λ_ka_k,j = 0 (25)
where λ₁, λ₂, . . .,λ_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 λ is not null - and putting μ₁ = −λ₁/λ, μ₂ = − λ₂/λ, . . ., μ_k = − λ_k/λ, one gets a_i,j = μ₁a_1,j + μ₂a_2,j + . . . +μ_ka_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.

I think that assuming that coefficients λ₁, λ₂, . . ., λ_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 wonder 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 λ's depend only on the row to which they belong and that , as a consequence, if they belong to the same row, even if the column varies, they are equal. In fact, if they - generally - depended on (i,j) - that is if they were λ_i,j as I would have expected them to be and not simply λ_i -, the theorem would not be valid any longer.

If one develops the determinant (24 ): $$\begin{vmatrix}a_{1,1}&\ldots&a_{1,k}&a_{1,j}\\\ldots&\ldots&\ldots&\ldots\\a_{k,1}&\ldots&a_{k,k}&a_{k,j}\\a_{i,1}&\ldots&a_{i,k}&a_{i,j}\end{vmatrix} = \lambda a_{i,j} + \begin{vmatrix}a_{2,1}&\ldots&a_{2,k}\\a_{k,1}&\dots&a_{k,k}\\a_{i,1}&\ldots&a_{i,k}\end{vmatrix} a_{1,j} + . . . \tag{24-25} $$, one can easily 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. In fact, if one calculates λ'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 sound demonstration should grant.

In fact, it is not sufficient to only prove that λ's are not dependent on j's, but it should be proved that they are equal when calculated on different columns of the matrix.

Am I right or not or what is totally escaping me in this demonstration ?

 

[fresh_42]

"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".

I don't think this is a good way to define it. At least I would speak of existing k rows or columns to avoid the misunderstanding of the first k. And the fact that the other row vectors are linear dependent, means that k is the maximal number with this property.

I like to think of the rank of a matrix as the dimension of the image space, i.e. if a matrix $A$ is representing the linear mapping $A\, : \,\mathbb{F}^n \longrightarrow \mathbb{F}^m$, then $\operatorname{rk}A = \dim A(\mathbb{F}^n)$.

As a formula, your definition can be written as: Let $A =[a_1, \ldots , a_n]$ (with row vectors $a_i$), then
$$
\operatorname{rk} A = \operatorname{max}\{k \in \mathbb{N}\,\vert \,\exists_{\pi \in \mathcal{Sym}(n)} \; \{a_{\pi(1)}, \ldots , a_{\pi(k)} \} \text{ is linear independent }\}
$$
and with mine
$$
\operatorname{rk} A = \dim \operatorname{span}\{a_1, \ldots , a_n\}
$$

 

[Conrad Manfried]

First of all, thanks a lot for answering. I agree with what you wrote. Maybe I did not manage to be clear: the definition I quoted is not mine, but it was taken from the book suggested by my teacher. Actually, the real purpose of my question was to know whether what I exactly copied from the book can be considered an effective demonstration - and I do not think so, but I may be wrong - or should deemed of as no more than a skilful "trick" based on indexes - which I strongly tend to think -.
If possible, this is what I would like to be told, because I would like to know whether I am actually loosing my time studying on a book where most of so called "demonstrations" are not valid as such and cannot be considered more than a sort of verifications of already known properties. This way I am likely to pass my exams as my teachers expect and request from me, but I will never learn to reason and distinguish between real proofs and tricky games - which I evaluate immensely more -.
Thanks again

 

[fresh_42]

If possible, this is what I would like to be told, because I would like to know whether I am actually loosing my time studying on a book where most of so called "demonstrations" are not valid as such and cannot be considered more than a sort of verifications of already known properties.

You completely lost me between eq. 23 and eq. 24. Is it supposed that $\operatorname{rk} A = k$?It is important to be exact on what is known, stated, a given condition and what is intended to be shown. In the above it is not, in my opinion. What is the theorem? I can only see a definition. Maybe it's simply too late here, but I find it hard to read.

 

[Conrad Manfried]

Thanks again for answering! Yes, the book I am studying on - and the teacher, who follows that book - supposes that rk A = k . What I exactly copied from the book - because I am not certain that a real demonstration can be performed this way - are intended to be the theorem's thesis and related demonstration.

The thesis - in the book - is:

"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".

The demonstration - at least, according to the book - is what follows in the text up to the final sentence:
"The theorem is thus proved as for rows and can be analogously demonstrated as for columns."

This is what is actually written in the book and I do not have but "to learn" and repeat it like an automa, but I am sure that this one cannot be considered a demonstration because it does not prove that λ's should be the same independently on the column on which they are calculated whilst - on the contrary - that would be absolutely necessary in order to be able to say that rows beyond k are linearly dependent - or am I wrong ? -.

Thanks

 

[Conrad Manfried]

Obviously, the book's "demonstration" - I did not include in the previous message - is exactly the one I reported in my first message:

"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:
λa_i,j+λ₁a_1,j+λ₂a_2,j+ . . . + λ_ka_k,j = 0 (25)
where λ₁, λ₂, . . .,λ_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 λ is not null - and putting μ₁ = −λ₁/λ, μ₂ = − λ₂/λ, . . ., μ_k = − λ_k/λ, one gets a_i,j = μ₁a_1,j + μ₂a_2,j + . . . +μ_ka_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."