Two Linear Algebra problems of completely different flavours

Eyy, what's up - what's down? I have my final high school exams this week, and, after that, I'll be free to study anything I want. I already started looking into tensor products and metric spaces (I know, kinda unrelated), and I really hope I can upgrade the level of Maths on this blog by the end of summer. I really wish to learn more advanced Maths topics, so I could also do some neat Physics.

Anyways, before my final exams, I thought that I should post something fun, in case I don't make it lmao. So here are my picks for this article! Of course, they are linear algebra problems, but I promise I will soon change the topic. As much as I'd love linear algebra (it's my favourite part of Mathematics at the moment), this blog is called 'Cris's Science Blog', not 'Cris's Linear Algebra Blog'.

The first problem comes from an Excellence Test from Timisoara. I was searching through some old Mathematical magazines of mine, and I found a remarkable number from the RMT (Revista Matematica Timisoara, the Mathematical Magazine of Timisoara), the one made for the 2017 Maths Olympiad. Of course, it brought back memories, since that was my brightest year. That particular number was dedicated to the memory of one of the greatest Mathematics teachers Romania ever had, Gheorghe Eckstein.

This problem looks like an X and O with matrices. Suppose you have a \(2n * 2n\) matrix, that is 'empty' at first. Two players, A and B, take successive  turns at writing numbers in the matrix. Once a number was written, it can't be erased. A wins if the determinant of the matrix isn't 0, while B wins if the determinant is 0. The question asks if there exists a strategy for which B always wins, no matter what A does.

Solution: After playing with my imaginary friends for a while, I figured out that there is, indeed, a strategy. Since the matrix doesn't involve any known numbers, it was safe to assume that the strategy consisted of doing something with the lines of the matrix.

Consider the first two lines of the matrix. If A puts a number \(x\) on the first line, B should put the same number right under it. By repeating this process, there will be two equal lines. Since the number \(2n\) is even, B will always be the one ending the process, and, thus, he will be in control of the determinant. It doesn't matter what A puts in the other lines; if A puts a number on one of the other lines, B can randomly put a number there, too. Eventually, A will be the one writing a number on one of the first lines, and B will duplicate that number.

The problem wasn't exactly hard, but it was a fun game. That's why I added the 'Intriguing Puzzles' tag.

Now comes a more scientific one. This one is taken from Evan Chen's 'Napkin'. I started studying University Maths from there, and I really recommend checking it out: http://web.evanchen.cc/napkin.html

Let V be a finite dimensional vector space, and let \(T: V->V\) be a linear map, and let \(T^n:V->V\) denote T applied n times. Prove that there exists an integer N such that \(V=Ker(T^N)⊕Im(T^N)\).

Solution: In order to talk about a direct sum, I want to prove that there is an N after which \(Ker(T^N)⋂Im(T^N)=Φ\). It is easy to see that, as N increases, the Kernel of the linear map increases, while the dimension decreases. Since V is finite dimensional, the dimension of the Kernel reaches a maximum, let's call it l. I want to prove that, after it reaches this maximum, the Kernel and the Image become distinct. Since the dimension of the Kernel remains the same, it is obvious that the Kernel will remain the same after iterations. Let N be the iteration of T for which the Kernel remains constant afterwards.

Let \(a\) be in both \(Ker(T^{N+1})\) and \(Im(T^{N+1})\). I want to prove that \(a=0\). We know that \(T^N(a)=O\), and, since a is in the image, we have that \(T^{2N+1}(b)=0\), for b, the pre-image of a. Thus, we know that b is in \(Ker(T^{2N+1})=Ker(T^{N})\). But, since \(T^{N+1}(b)=a\), we get that \(a=0\), exactly as we wanted, so we can talk about a direct sum.

Now, we know that \(dim (Ker(T^N) ⊕ Im(T^N)) = dim Ker(T^N) + dim Im(T^N) = dim V\), and, since the Kernel and the image are subspaces of V, we get that their direct sum is exactly V, as we wanted.

I found this problem rather interesting, and, from what I saw, it works as a lemma. I haven't tried writing this as actual vector spaces instead of abstract notations, but I might look into it. Until then, I have to study Literature for my exam on Monday. Wish me luck, and more posts are coming! At first, they'll still be Linear Algebra, but after a while, trust me, they'll become more diverse. Tune in next time for something completely random about matrices!

Cris.