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Recall the definition of the rank of a matroid $(V, \mathcal{I})$:

$$ r(A) = \operatorname{rank}(A) = \max_{I \in \mathcal{I}}\{ | A \cap I | \} = \max\{ |I| : I \subseteq A, I \in \mathcal{I} \}$$

I was trying to prove that the rank of a matroid is a sub-modular function, i.e. that the following inequality holds for all subsets of the ground set (i.e. $\forall A, B \subseteq V$):

$$ r(A) + r(B) \geq r(A \cup B) + r(A \cap B)$$

I tried a "picture proof" by drawing a couple of sets and seeing how their intersection with independent elements behaved and I can only conclude that its in fact an equality. I am sure there is something wrong with that method and its not a real proof but wasn't sure how else to approach it. Does someone have a proof or a suggestion on good direction I might try to actually prove this result?

Also, is this suppose to be "intuitively obvious"? Because its not completely obvious for me.

Andrés E. Caicedo
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Charlie Parker
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3 Answers3

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Start with $I_1$ maximal independent in $A\cap B$. We have $r(A\cap B)=|I_1|$. By the property of independent sets, we may extend $I_1$ to $I$, maximal independent in $A\cup B$. We have $r(A\cup B)=|I|$. Now, partition $I=I_2\cup I_3\cup I_4$, where $I_2\in A\setminus B$, $I_3\in B\setminus A$, $I_4\in A\cap B$. By construction, $I_1\subseteq I_4$; however by the maximality of $I_1$, in fact $I_1=I_4$. Hence $r(A\cup B)=|I_1|+|I_2|+|I_3|$. This takes care of the RHS.

Now, $I_1\cup I_2$ is a subset of $I$, hence independent, and contained entirely in $A$. It might not be maximal independent in $A$, so all we can say is $r(A)\ge |I_1|+|I_2|$. By a similar argument $r(B)\ge |I_1|+|I_3|$.

Putting it all together, we get $$r(A)+r(B)\ge 2|I_1|+|I_2|+|I_3|=r(A\cup B)+r(A\cap B)$$

A bit more about the strictness. Suppose our ground set is the edges of a triangle graph (three vertices, three edges called $a,b,c$), with all cycle-free sets independent. That is, all subsets are independent, except all three edges together.

Set $A=\{a,b\}$, $B=\{a,c\}$. $r(A)=r(B)=2$, so LHS=$2+2=4$. $r(A\cap B)=1$, since $A\cap B=\{a\}$. However, $r(A\cup B)=2$, since $A\cup B=\{a,b,c\}$ yet all independent sets are of size 2 or less. Hence RHS=$2+1=3$.

vadim123
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  • There is one crucial part I don't think I understand. How come $I_1 = I_4$? I am not quite sure how to express my doubt but I just can't see why that is necessarily true. – Charlie Parker Mar 16 '15 at 03:24
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    If $|I_1|<|I_4|$ then $I_1$ is not maximal cardinality independent in $A\cap B$. – vadim123 Mar 16 '15 at 03:27
  • That explains my confusion, that wasn't what I was confused about, I just realized what I don't understand is why $I_1 \subseteq I_4$ is necessarily true. – Charlie Parker Mar 16 '15 at 03:29
  • I guess the part that confuses me is that, we may extent $I_1$ with some element x in A or B (to try to form some independent set in $A \cup B$), however, the union $I_1 \cup {x}$ is not necessarily an independent set in the matroid, right? not sure if that matters. – Charlie Parker Mar 16 '15 at 03:41
  • For me what seems correct is the following. Since $I_4 \subseteq A \cap B$ then, its size must be at most the size of the maximal independent subset in $A \cap B$. Hence, the correct thing to say for me is $|I_4| \leq |I_1|$. That is obviously true. However, the opposite is not obvious to me. i.e. is $|I_1| \leq |I_4|$ true? – Charlie Parker Mar 16 '15 at 03:49
  • Oh I see, since both $I_1$ and $I_4$ are proper subsets of $A \cap B$ and since $I_1$ is maximal, if $I_4$ is maximal, then it just means that $I_1$ wasn't actually maximal and that the max is actually $I_1$. I think I got that part now...? Thnx btw. :) – Charlie Parker Mar 16 '15 at 03:52
  • actually not sure if I got it... – Charlie Parker Mar 16 '15 at 04:49
  • I guess my key question would be, I don't think we can construct max $A \cup B$ from $I_1$ because, it might be possible that, by reducing the size of the elements we choose from the intersection $A \cap B$ and instead, choose elements from $A$ or $B$, that we increase the size of the independent set from $A \cup B$ so Im not convinced $I_4 = I_1$ necessarily. – Charlie Parker Mar 16 '15 at 04:55
  • All of your questions boil down to understanding the independence property of matroids. Given set $S$ and an independent subset $I$, either $I$ is maximal, or there is a larger set $J$. Then, you can add elements to $I$ (drawn from $J$), until $|I\cup {j_1}\cup {j_2}\cup\cdots\cup {j_t}|=|J|$. – vadim123 Mar 16 '15 at 05:00
  • Thanks for your help, if you don't mind adding just a few more details to "By the property of independent sets, we may extend $I_1$ to $I$, maximal independent in $A \cup B$", which seems to be the part of the proof I am a little shaky on. I do understand what you said above but not sure if I know how to fill in the details. I know is probably annoying because you understand it much better and i apologize. But I do appreciate your time and efforts :) – Charlie Parker Mar 16 '15 at 14:29
  • Or let me try to write an answer according to what I understand and you can correct it if its wrong. – Charlie Parker Mar 16 '15 at 14:37
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I am writing an answer to make sure I understand vadim's answer.

proof:

let $I_1 = \max_{I \in \mathcal{I}}\{|(A \cap B) \cap I|\}$, i.e., the maximal independent set of $A \cap B$. Recall the property of independent sets in matroids; if $|I| > |J|$ then there exists a sequence of elements $x_i \in I \setminus J$ such that at some point $|I| = |J \cup \{ x_1\} \cup ... \cup \{ x_k \}|$.

Now lets consider $I$, the maximal independent set of $A \cup B$. Obviously the size of $I$ cannot be less that $I_1$. i.e. $|I_1| \leq |I|$ (if the opposite where true, then clearly $I$ wouldn't be a maximally independent set of $A \cup B$!). Anyway, in the case where the inequality is strict $|I_1| < |I|$, then there exists one or more elements in $I$ not in $I_1$ that we can add to $I_1$ until $I_1$ matches in size to $I$. i.e. $|I| = |I_1 \cup \{ x_1\} ... \{ x_k \}|$.

Once that last equality is true then we can reason about where the elements $x_i$ came from.

Clearly, none of the $x_i$'s came from the intersection $A \cap B$, since if that were true, then $I_1$ wouldn't be maximal. Hence, they came from $A \setminus B$ or $B \setminus A$. Hence we have:

$$ I_1 \cup \{ x_1\} ... \{ x_k \} = I_1 \cup I_2 \cup I_3 $$

where $I_2 \subseteq A \setminus B$ and $I_3 \subseteq B \setminus A$.

Now $I_1 \cup I_2 \subseteq A$, however it might not be maximal. Hence $r(A) \geq |I_1| + |I_2|$. Similarly for $I_1 \cup I_3 \subseteq B$ we have $r(B) \geq |I_1| + |I_3|$.

Hence we get:

$r(A) + r(B) \geq |I_1| + |I_2| + |I_1| + |I_3|$

However, notice that $r(A \cup B) = |I| = |I_1 \cup I_2 \cup I_3| = |I_1| + |I_2| + |I_3|$ and $r(A \cap B) = |I_1|$. Hence:

$$r(A \cup B) + r(A \cap B) = |I_2| + 2|I_1| + |I_3|$$

Which Concludes the proof:

$r(A) + r(B) \geq |I_1| + |I_2| + |I_1| + |I_3| = r(A \cup B) + r(A \cap B)$

Sidak
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Charlie Parker
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For a quick counterexample to equality, let $B$ be a basis and suppose there's an $e$ not in $B$ such that $C(e,B)$ is the the $B$-fundamental circuit of $e$. Then:

  • $r(B) = r(M)$,
  • $r(e) = 1$,
  • $r(B \cup e) = r(M)$, and
  • $r(B \cap e) = r(\emptyset) = 0$.
NoName
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