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I have a somewhat unconventional question, and I apologize if it does not adhere to the criteria for inquiries on this website. However, as I have received this task on relatively short notice, I urgently need help from someone.

I have to prepare a presentation on the topic of "Diagonalizability" by Thursday, and in addition to the theory, I am supposed to discuss problems from mathematics competitions where one can apply the theory or where the theory is helpful in solving the problems (a didactic approach, so to speak).

The issue I'm facing is that I have no idea where to find problems from mathematics competitions related to the topic of "Diagonalizability." I wanted to ask if someone here could help me get such problems or guide me to a place where I can find them. Is there a website where I can somehow "filter" problems from mathematics competitions based on keywords, so I can find the ones I need?

I appreciate any kind of help!

MathGeek
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    Presumably, "diagonalisability" is not part of the problems' statements, but of their solution. The most simple application is that for some functions $f$ of matrices, computing $f(M)$ is hard in general, but easy if $M$ is diagonal or diagonalisable.. For instance, computing $\exp(M)$ using formula $$\exp(M) = \sum_{k=0}^{+\infty} \frac{1}{k!}M^k$$ is a lot of work if $M$ is not diagonal, but if $M$ is diagonal then you just have to take the real-number exponentials of the diagonal coefficients; and if $M = P^{-1} D P$ then $\exp(M) = P^{-1} \exp(D) P$. – Stef Jan 15 '24 at 13:02
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    This is also true for computing $M^k$ for a large $k$, although computing $M^k$ is easier than $\exp{M}$. For instance consider the problem "Compute the 100th Fibonacci number". You know the recurrence formula $\operatorname{fib}(n+2) = \operatorname{fib}(n+1) + \operatorname{fib}(n)$, but applying this formula 100 times is tedious. Instead, consider $$F_n = \left( \begin{matrix} \operatorname{fib}(n+1)\ \operatorname{fib}(n) \end{matrix} \right)$$ and $$M = \left( \begin{matrix} 1 & 1 \ 1 & 0 \end{matrix} \right) $$ and see how $F_{n+1} = M F_n$ and $F_{100} = M^{100} F_0$. – Stef Jan 15 '24 at 13:09
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    Then you can diagonalise $M$ during the presentation, and use the diagonal form to compute $M^{100}$ in front of their eyes as if computing powers of matrices was easy for you. (Spoiler: the diagonal coefficients are going to be the golden ratio and its inverse) – Stef Jan 15 '24 at 13:12
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    To find past competitions, try https://artofproblemsolving.com/community Maybe try "diagonalizability" in their search box. – GEdgar Jan 15 '24 at 13:45

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