This has been a long time coming. I have awakened from my long slumber of procrastination, coursework and computer games.
What a wonderful time to be alive, amirite? The best time to go outside, party with friends... Oh wait. Yup people, stay inside. As much as it pains everyone, in order to ease the pressure that has been put on the medical system, we have to stay inside and avoid getting the virus. Don't be overly-terrified, as the chances of the virus being fatal are small for most of us, but don't be ignorant; COVID-19 is an aggressive virus.
Since I've been staying indoors in self-isolation for two weeks now, I started looking through my old stash of Math results, so I came across the Cristi Endomorphisms. It was either this or a result in Probability and Statistics.
The problem, which was posed at the Traian Lalescu contest at the Polytechnic University of Bucharest in December, goes as follows:
Let V be a vector space and T and S two of its endomorphisms such that \(V=Ker(T)+Ker(S)=Im(T)+Im(S)\).
a) If V has finite dimension, prove that the two are direct sums.
b) Does the result still apply if V has infinite dimension?
Solution:
Part a) is pretty straightforward. We know the following: \(dim(U+W)=dim(U)+dim(W)-dim(U∩W)\).
By applying this on our current problem, we get that
\(dim(V)=n=dim(Ker(T))+dim(Ker(S))-dim(Ker(S)∩Ker(T))=\)
\(=dim(Im(T))+dim(Im(S))-dim(Im(S)∩Im(T))\)
Sum up the two equalities, and we get that \(2n=dim(Ker(T))+dim(Im(T))+dim(Ker(S))+dim(Im(S))-\)
\(-dim(Ker(T)∩Ker(S))-dim(Im(T)∩Im(S))\)
By using the rank-nullity theorem, we get that \(dim(Ker(T)∩Ker(S))+dim(Im(T)∩Im(S))=0\), so both dimensions are 0. Thus, \(Ker(T)+Ker(S)=Ker(T)⊕Ker(S), Im(T)+Im(S)=Im(T)⊕Im(S)\).
The second part is a bit tricky. It asks for either a proof or a counterexample. By now you probably figured out that it wants a counterexample; that's how 90% of these problems work. Now, what infinitely dimensional vector spaces do we know? No, not that space of matrices of infinite order. \(R/Q\) of course!
Let us consider \(U={U_1, U_2, ...}\) a countable subdivision of a Hamel basis. For the rest of the solution, we shall deal with this subdivision. The elements of the basis outside of this subdivisions shall be considered fixed under the endomorphisms' actions.
What we want to find out now is how this countable subdivision should behave according to the endomorphisms. Let \(S(U_1)=S(U_2)=U_1, S(U_3)=0, S(U_4)=U_2, ...S(U_n)=S(U_{n-2})\). Why do we have to do this? So that the Kernel of S is generated by the elements \(U_1, U_2, U_3\), while the Image is still the whole V. The Kernel of S is basically \(a(U_1-U_2)+bU_3\).
Now for T: Choose \(T(U_1)=U_1, T(U_2)=0, T(U_3)=U_3\), and the rest are 0. This way, the Kernel of T is generated by \(U_2\) and all other elements of the Hamel basis other than \(U_1\) and \(U_3\). Thus, the sum of the Kernels is V, the sum of the images is V, and yet the images don't form a direct sum. If you want to upgrade this, so that the Kernels don't form a direct sum either, let \(S(U_4)=0\) and then continue with \(S(U_n)=U_{n-3}\) instead of \(n-2\). Thus, the Kernels intersect in a subspace generated by \(U_4\).
I really enjoyed this problem, as it challenges the student's understanding of vector spaces. Most problems involve a finitely dimensional space, or are easily translated into matrices, but this problem requires some out-of-the-box thinking, and I felt really pleased with myself for solving it. I admit, I was so pleased with it, that I started telling literally everyone about the Cristi Endomorphisms. I'm certain that most of my friends can agree that I annoyed them a bit with this problem (;
See you next time with that probability result, some complex analysis that I found, yet another matrix problem, or something completely random! Until then, stay safe people!
Cris
