2
$\begingroup$

Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3), (1, 4), (3, 3), (3, 4),$ and $(5, 4)$ are changed to obtain another $7×6$ real matrix $N$. Suppose that the rank of $N$ is 4. What could be the rank of $M$?

I want to list all the possibilities. I can see that it should be less than $6$ and the rank is less than equal to the minimum of no. of.rows and cols. I did some examples, it seems like 2,3,4 are possible. But I am not getting a general idea. Kindly help me with this. Thank you.

I am preparing for competitive exams and this is a question from one such paper.

$\endgroup$
1
  • $\begingroup$ It also seems to me that all ranks between 2 and 6 are possible. A proof should state a) that the rank of M cannot be 0 or 1 and b) give examples of rank 2,3,4,5 and 6. It should be possible to give examples having entries 0 and 1 only. $\endgroup$
    – Helmut
    Mar 31, 2020 at 12:35

2 Answers 2

2
$\begingroup$

The rank of a matrix is determined by the dimension of the span of the row vectors or the span of the columns vectors. Seeing the matrix as a list of row vectors then the condition of the problem say that you can change at most three row vectors (row one, three and five) , similarly seeing the matrix as a list of row columns the condition of the problem say that you can change at most two column vectors (column three and four).

Then you must see how much you can downgrade or upgrade the rank of the matrix studying how much you can make these row or column vectors dependent or independent.

At most, without any analysis, you can diminish or increase the rank of the matrix by two, because it is enough to study just the column vectors (because the rank its the same as studying the row vectors), then it is enough to show that this could be possible giving some example of a matrix of rank four that, when changing the values of the coefficients of the exercise, it down to a matrix of rank two (this example is easy to find using the canonical column vectors that have zeros at all places except at a position) or it increases it rank to six.

$\endgroup$
1
$\begingroup$

Since \begin{aligned} \operatorname{rank}(M)&=\operatorname{rank}(M-N+N)\le\operatorname{rank}(M-N)+\operatorname{rank}(N)\\ \operatorname{rank}(N)&=\operatorname{rank}(N-M+M)\le\operatorname{rank}(N-M)+\operatorname{rank}(M), \end{aligned} we have $$ \operatorname{rank}(N)-\operatorname{rank}(N-M) \le\operatorname{rank}(M) \le\operatorname{rank}(M-N)+\operatorname{rank}(N). $$ By assumption, $\operatorname{rank}(N)=4$. Also, since all changes from $M$ to $N$ occur on two columns (namely, columns $3$ and $5$), we have $\operatorname{rank}(N-M)=\operatorname{rank}(M-N)\le2$. Therefore $2\le\operatorname{rank}(M)\le6$. To show that $\operatorname{rank}(M)$ can be any integer from $2$ to $6$, you need to exhibit some concrete examples. Below are some examples of changes of $\{0,1\}$-matrices that work over any field (not just the reals). Note that we are not changing only some of the entries $m_{13},m_{14},m_{33},m_{34},m_{54}$, but all of them.

Rank changes from $2$ to $4$: $$ \pmatrix{ 0&0&\color{red}{0}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{0}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{0}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1} \to\pmatrix{ 0&0&\color{red}{1}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{1}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{1}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1}. $$ Rank changes from $3$ to $4$: $$ \pmatrix{ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{0}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1} \to\pmatrix{ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{1}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1}. $$ Rank changes from $4$ to $4$: $$ \pmatrix{ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{0}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1} \to\pmatrix{ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{1}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1}. $$ Rank changes from $5$ to $4$: $$ \pmatrix{ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&1&0&0&0&0\\ 0&0&\color{red}{1}&\color{red}{0}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{1}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1} \to\pmatrix{ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&1&0&0&0&0\\ 0&0&\color{red}{0}&\color{red}{1}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&\color{red}{0}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1}. $$ Rank changes from $6$ to $4$: $$ \pmatrix{ 0&0&\color{red}{1}&\color{red}{1}&0&0\\ 0&1&0&0&0&0\\ 0&0&\color{red}{1}&\color{red}{1}&0&0\\ 1&0&0&0&0&0\\ 0&0&0&\color{red}{1}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1} \to\pmatrix{ 0&0&\color{red}{0}&\color{red}{0}&0&0\\ 0&1&0&0&0&0\\ 0&0&\color{red}{0}&\color{red}{0}&0&0\\ 1&0&0&0&0&0\\ 0&0&0&\color{red}{0}&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1}. $$

$\endgroup$
1
  • $\begingroup$ Thank you for the wonderful answer. They have used all the entries $m_{13},m_{14},m_{33},m_{34},m_{54}$. So the person who set this question seems know about this and what could be his intuition about this problem? Kindly share your thoughts. $\endgroup$ Mar 31, 2020 at 16:06

Not the answer you're looking for? Browse other questions tagged .