Does rank additivity still hold after removing one matrix from a rank decomposition?

Key takeaway: Rank identities can look stable at first glance, but they often break the moment you delete one summand. The key is whether the remaining matrices still behave independently.

Yes, it is necessarily true.

We are given

$A_1+\cdots+A_k=I_n$

and

$n=\operatorname{rank}(A_1+\cdots+A_k)=\operatorname{rank}(A_1)+\cdots+\operatorname{rank}(A_k).$

Since $\operatorname{rank}(I_n)=n$, the equality

$\operatorname{rank}(A_1+\cdots+A_k)=\sum_{i=1}^k \operatorname{rank}(A_i)$

forces equality in the rank subadditivity chain. In particular, for the sum of the last $k-1$ matrices we have

$A_2+\cdots+A_k=I_n-A_1.$

Now apply rank subadditivity:

$\operatorname{rank}(I_n)=\operatorname{rank}(A_1+(A_2+\cdots+A_k))\le \operatorname{rank}(A_1)+\operatorname{rank}(A_2+\cdots+A_k).$

Because the total rank already equals the sum of the individual ranks, the only way the full equality can hold is if the remaining block also satisfies

$\operatorname{rank}(A_2+\cdots+A_k)=\operatorname{rank}(A_2)+\cdots+\operatorname{rank}(A_k).$

So the answer is yes.

A compact way to view it is that the original equality leaves no room for hidden overlap among the ranges of the summands; removing one term preserves the same structure for the rest.

---

Pitfalls the pros know 👇 The main trap is treating rank like ordinary addition. You cannot write rank of a sum as sum of ranks unless special equality conditions are met. Another mistake is forgetting that the hypothesis already encodes a very strong extremal case.

What if the problem changes? If the first equality were weakened to only $A_1+\cdots+A_k=I_n$ without rank additivity, then the conclusion could fail. Rank additivity for the full sum is the decisive hypothesis here.

Tags: matrix rank, rank subadditivity, identity matrix