A submatrix $B$ consisting of "s" rows of $A$ is selected from an n-square matrix $A$ of rank $r_{A}$. prove that the rank of $B$ is equal to or greater than $r_{A}+s-n$.

My thoughts:

I start with an easy case, says $A=I_4$. Then, by selecting 2 first rows of $A$. We obtain a matrix $B$:$$\begin{pmatrix} 1 &0 &0 &0 \\ 0&1 &0 &0 \end{pmatrix}$$ So, the rank of $B=4+2-4=2$. At least, I know the statement is true for this trivial situation. But can anyone help me to figure out the general situation about the problem? Thanks in advance.


2 Answers 2


Hint: Note the following: $r_A\leq n$ and $r_B\leq s$. With this in mind, instead of trying to prove that $r_B\ge r_A+s-n$ try to prove an equivalent inequality that has $r_B, r_A$ one one side and $n,s$ on the other.

As was pointed out below, my hint has issues and I don't see a way to salvage it. I am unable to delete this answer, so, for self-containment, I'm copying Leon Sot's answer:

Rearrange to get $$r_A-r_B\leq n-s. $$ So the difference in rank is less than the difference in rows. There are $r_A-r_B$ rows that add the the rank of $A$ but not to the rank of $B$. These rows cannot be in the $s$ selected rows since then this would increase the rank of $B$. Therefore, they are one of the $n-s$ rows not in $B$. The inequality follows.

  • $\begingroup$ Thank you for your hint so much. It helps me to work it out. =) $\endgroup$
    – nam
    Apr 11, 2014 at 16:31
  • $\begingroup$ @nam I'm glad.${}$ $\endgroup$
    – Git Gud
    Apr 11, 2014 at 16:31
  • $\begingroup$ Kindly provide hint 2. $\endgroup$ Nov 24, 2019 at 7:20
  • $\begingroup$ @ChandramauliChakraborty Subtract the two first inequalities in my answer. $\endgroup$
    – Git Gud
    Nov 24, 2019 at 21:32
  • $\begingroup$ @GitGud, then I would get $r_A-s \leq n-r_B$. $\endgroup$ Nov 25, 2019 at 13:42

Here $A$ is $n\times n$ matrix and $r_A$ be the rank of $A$. Now we need to eliminate $n-s$ rows from $n$ rows. Let $C=\{A_{i_k*}:1\leq k\leq r_A\} $, be the set of basis of row space of A , i.e , $\mathcal{R}(A)$, where $A_{i_k*}$ denote the $i_k\textit{-th}$ row of the matrix A.

Now choose any row $A_{j_m*}$, there would be two possibilities, i.e. either $A_{j_m*}\in C$ or $A_{j_m*}\not\in C$.If $A_{j_m*}\in C$, then the new matrix is of order $(n-1)\times n$ has rank $r_A-1$, while in the other case, $r_B=r_A$, since $A_{j_m*}\in span(C)$.

Hence if we want the minimum of $r_B$, then it should be the worst case where all the $n-s$ rows belong to $C$. Hence $r_B\geq r_A-(n-s)$.


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