What properties does $A\in M_n(\Bbb F),n\geqslant2\ \ $ have if $\operatorname{rank}(A)=\det(A)$?

My work:

$A$ is either $\text{regular}$ or a $\text{zero-matrix}$.

~For non-trivial case, let $C\in M_n,n\geqslant2$ be regular and $\det(C)=n$ $$d(n):=\{k\in\Bbb N:k\mid n\}$$

Depending on whether $n$ is $\text{prime}$ or $\text{composite}$, we multiplied some number($\leqslant d(k)$) of columns(rows) by some $k$.

To visualise (the simplest LaPlace development or evaluation of a diagonal matrix determinant), let: $$B=\begin{bmatrix}I_{n-1}&0\\0&n\end{bmatrix}$$ $\prod_{i=1}^n a_{ii}=\det(B)=n\cdot\det(I_{n-1})\in \Bbb N$

For $n\in\Bbb N_{2n+1}$ it would work even for $-I_{n-1}$

Is $B\sim C$ s.t. the equivalence realized by an arbitrary number of the $3^{\text{rd}}$ type of elementary transformations and/or an even number of the first two types? Then the matrices $B\in[C]$? in order to $|B|=|C|,\ \ B,C\in M_n(\Bbb Z)$

  • $\begingroup$ I cannot understand what you mean by $r(A)$ (is it the rank of the matrix?), and by "regular matrix" (is it maybe a non-singular matrix?). Then, I dont' understand why you pick any "regular matrix", and state that its determinant is necessarily an integer (it could be any element of the field $\mathbb F$). Then I don't understand what "multiply $d(k)$ columns", if $d(k)$ denotes the set of divisors of $k$ (for some $k$?). As it is stated, this question is unclear. $\endgroup$ – Crostul Jan 15 at 17:43
  • $\begingroup$ $r(A)\iff\operatorname{rank} (A)$ Regular matrix $B\in M_{mn}(\mathbb F)$ has a full rank $r(B)=\min\{m,n\}$ $A\in M_n(\mathbb F)$ is a square matrix and is regular. It is necessarily a positive integer because $r(A)\in\mathbb N$ We can multiply, if we want so, by distinct divisors of $n$, which isn't important. $\endgroup$ – Invisible Jan 15 at 17:53

Such a matrix can be simply build in the following way.

Let $A$ be any non singular square $n \times n$ matrix, over any field $\Bbb F$ extending the rationals ($\Bbb Q \subseteq \Bbb F$).

We have $\det A \neq 0$, while the rank of $A$ is $n$. Let $a_1, \dots , a_n \in \Bbb F$ such that $$a_1 \cdots a_n = n (\det A)^{-1}$$ Then if you multiply for all $j \in \{ 1, \dots , n \}$ the $j$-th row/column of $A$ by $a_j$, you will get a new matrix $\tilde{A}$ whose determinant is $n= r(\tilde{A})$.

In particular, you don't need that the matrix has integer entries, nor integer eigenvalues. In other words, it could be (almost) anything.

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  • $\begingroup$ Thank you very much, I agree with that. $\endgroup$ – Invisible Jan 15 at 18:10

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