Gheorghe Țițeica 2025 Grade 11 P2

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Source: Gheorghe Țițeica 2025

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#1 h Mar 28, 2025, 9:42 PM • 1 Y

Y by MS_asdfgzxcvb

Let $n\geq 2$ and $A,B\in\mathcal{M}_n(\mathbb{C})$ such that $\{\text{rank}(A^k)\mid k\geq 1\}=\{\text{rank}(B^k)\mid k\geq 1\}.$ Prove that $\text{rank}(A^k)=\text{rank}(B^k)$ for all $k\geq 1$ .

Cristi Săvescu

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#2 h Mar 30, 2025, 9:39 AM • 1 Y

Y by MS_asdfgzxcvb

Very cute problem.

For any matrix $J \in \mathcal{M}_n(\mathbb{C})$ the sequence $\left(\text{rank}(J^k)\right)_{k \geq 1}$ is strictly decreasing up to an index, then constant. This can be easily proved using the Jordan form of $J$ , or even in an elementary way, by Frobenius. From now I will use $R_J$ for the set $\{\text{rank}(J^k)\mid k\geq 1\}$ .

Now we basically only need to observe two things:

The first thing is: say $k_A, k_B$ are those indices from where the rank-sequence of $A$ , respectively $B$ , get constant. Then $k_A=k_B$ and $\text{rank}(A^{k_A})=\text{rank}(B^{k_B})$ . The first part is because $|R_A|=k_A$ and $|R_B|=k_B$ . The second part is because $\text{rank}(A^{k_A})$ and $\text{rank}(B^{k_B})$ are the smallest terms in the sets $R_A$ and $R_B$ , respectively.

The second thing is to just observe the following equality of sets: $\{\text{rank}(A^1) > \text{rank}(A^2) > \ldots > \text{rank}(A^{k})\}=\{\text{rank}(B^1) > \text{rank}(B^2) > \ldots > \text{rank}(B^{k})\}$ , where $k=k_A=k_B$ . The result is now clear; just match the terms from biggest to smallest (one can even do this by induction, but even that would be kind of an overkill, as the number of steps we have to do is finite).

This post has been edited 2 times. Last edited by RobertRogo, Mar 30, 2025, 9:47 AM

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