Find matrix rank depending on two parameters with entries in $\mathbb{Z/7Z}$

Question

How do find the rank of the given matrix depending on parameters $a, b \in \mathbb{Z}_{7}$?

$$ A\in Mat_{3,4}(\mathbb{Z}_{7}), A=\begin{pmatrix} \overline{2} &b &\overline{1} &\overline{3} \\ \overline{-3} &\overline{3} b & \overline{2} a& \overline{0}\\ a & \overline{4} -a & \overline{-3} &\overline{1} +\overline{2} b \end{pmatrix}$$

I have found matrix rank by reducing matrix to echelon form, but there were no parameters involved in those questions, even more, two. I've spent days trying to figure out a solution by different methods (e.g by changing (-3) to 4 [they both give the same reminder when divided with 7] and then performed row operations, but it just did not worked out since i did not manage to reduce “a” to 0 etc. Then I started wondering whether I should just start by letting a/b equal to zero and watch different cases separately and reduce them to a desired form separately? I appreciate your help and explanations.

Answer 1

Here are some initial steps to get you started. For convenience, I'm omitting the bars.

Since we're only looking for the rank of this matrix, we can also apply column operations. $$ \pmatrix{2 &b &{1} &{3} \\ {-3} &{3} b & {2} a& {0}\\ a & {4} -a & {-3} &{1} +{2} b} \leadsto\\ \pmatrix{1 &4b &4 &5 \\ {-3} &{3} b & {2} a& {0}\\ a & {4} -a & {-3} &{1} +{2} b} \leadsto\\ \pmatrix{1 &4b &4 &5 \\ 0 & b & {2}a-2& 1\\ a & {4} -a & {-3} &{1} +{2} b} \leadsto\\ \pmatrix{1 &4b &4 &5 \\ 0 & b & {2}a-2& 1\\ 0 & {4} -a-4ab & -3-4a &2 +{2} b} \leadsto\\ \pmatrix{1 &5 &4 &4b \\ 0 & 1 & {2}a-2& b\\ 0 & 2+2b & -3-4a &4-a-4ab}.\phantom{\leadsto} $$ The first three steps were row operation, and the last was a permutation of the columns. Perhaps you can take it from here.

Answer 2

In fact, in such cases, when you have spent days on a question as is the case, consider (computed-assisted) "brute force", i.e., build a program that browses all cases, which are not that numerous ($7^2=49$ such matrices).

[All right, it's not considered as a proof]

See below a rather simple (Matlab) program which does the job and finds a single case where rank$(A)<3$ for

$$a=4 \ \ \ \text{and} \ \ \ b=5$$

giving matrix :

$$\begin{pmatrix}2&5&1&3\\4&1&1&0\\4&0&4&4\end{pmatrix}$$

More precisely, in this case, rank(A)=$2$ because the $2 \times 2$ upper left block has determinant $2 * 1 - 4 *5 = 3 \ \text{mod} \ 7 \ \ne 0$.

 for a=0:6
    for b=0:6 
       M=[2,b,1,3;
         -3,3*b,2*a,0;
          a,(4-a),-3,(1+2*b)];
       D=[det(M(:,[2,3,4]),det(M(:,1,3,4)),det(M(:,1,2,4),det(M:,[1,2,3])];
       D=mod(round(D),7);
       if not(any(D));% if all 3x3 det. are zeros
          [a,b]
       end;
    end;
 end;

Answer 3

Here is a rigorous answer.

I thought at first to group it with the previous one, but its nature is so different that I prefer to write a separate answer.

I will take the same conventions as Ben Grossmann : no overbars, and sign $\leadsto$ meaning "rank-equivalent" (having the same ranks depending in the same way on the values of $a$ and $b$).

Starting from Preliminary remark: as we can extract from

$$A=\pmatrix{2 &b &1 &3 \\ -3 &3 b & 2a& 0\\ a & (4-a) & -3 &(1+2b)},\tag{1}$$

a submatrix $\pmatrix{2 &3 \\ -3& 0}$ with determinant $5 \ne 0$, rank$(A) \ge 2$.

We are going to perform on $A$ different row (R) or column (C) operations that do not change the rank of the matrix:

As row number two, denoted $R_2$, can be written (with operations mod. $7$) :

$\pmatrix{4 &-4b &2a &0}$, in which we can factorize $2$ giving $\pmatrix{2 &-2b &a &0}$ ; subtracting this new row at the same time to row $R_1$, gives

$$\leadsto \ \ A':=\pmatrix{0 &3b &(1-a) &3 \\ 2 & -2b & a& 0\\ a & (4-a) & -3 &(1+2b)}\tag{2}$$

Let us now consider $C_1$ (column 1) written under the form (always with the rules mod. 7):

$$\leadsto \ \ \pmatrix{0 &3b &(1-a) &3 \\ 2 & -2b & a& 0\\ \color{red}{-6}a & (4-a) & -3 &(1+2b)}\tag{3}$$

We can factorize $2$, this time in column 1, and simultaneously, as we have done before when working on rows, subtract this new column 1 to column 3, yielding:

$$ \leadsto \ \ A''':=\pmatrix{0 &3b &(1-a) &3 \\ 1 & -2b & (a-1)& 0\\ -3a & (4-a) & -3(a-1) &(1+2b)}\tag{4}$$

If $a-1 \ne 0$ (we will come back on this hypothesis later), we can factorize it, and matrix $A'''$ is rank-equivalent to:

$$ \leadsto \ \ \ A'''':=\pmatrix{0 &3b &1 &3 \\ 1 & -2b & -1& 0\\ -3a & (4-a) & -3 &(1+2b)}\tag{5}$$

As a consequence, the rank of $A''''$ (and therefore the rank of $A$) is exactly equal to $2$ if and only if the four $3 \times 3$ determinants that one can extract from $A''''$ are all equal to $0$.

This gives the following system

$$\begin{cases}D_1&=&4a-4b-2-2b^2&=&0& \ (A)\\ D_2&=&-2a-2b-10&=&0& \ (B)\\ D_3&=&-3a-3b-18ab-6b^2&=&0& \ (C)\\ D_4&=&-a+2b+3ab+4&=&0& \ (D)\end{cases}$$

where $D_k$ is the determinant of the submatrix of $A''''$ obtained by suppressing the $k$th column.

The previous system can be simplified into:

$$\begin{cases}2a-2b-1-b^2&=&0& \ (A)\\ -a-b-5&=&0& \ (B)\\ -a-b-6ab-2b^2&=&0& \ (C)\\ -a+2b+4+3ab&=&0& \ (D)\end{cases}$$

Out of this system, one is able to build a system of 2 linear equations with two unknowns ; the first of these equations is (B), the second one being obtained by combination of the other equations $-3 \times (A)+(C)+2 \times (D)$.

This gives finally

$$a=4, \ \ b=5$$

But we have left a special case $a=1$ to be treated apart. Let us establish that this case cannot happen.

If we had not simplified (4) into (5), we would have had, by taking $a=1$ in (4):

$$A'''=\pmatrix{0 &3b &0 &3 \\ 1 & -2b & 0 & 0\\ -3 & 3 & 0 &(1+2b)}\tag{4}$$

Three of the four determinants extracted from this matrix are evidently $0$ but one of them (obtained by removing the third column) has value $b^2+2$.

But quadratic equation $b^2+2=0$ has no solution in $\mathbb{Z_7}$ ($-2$ is not a quadratic residue in this field), ruling out the possibility for $a$ to be equal to $1$. $\square$