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Mar 7, 2021 at 2:07 answer added ZeroTheHero timeline score: 2
Mar 7, 2021 at 1:13 comment added Cosmas Zachos Nothing deep: the kernel expands, becomes bigger, as the dimensionality of the matrix becomes bigger. The transposes of (1,....,-1), (0,1,...,-1,0), (0,0,1,...,-1,0,0) .... are the null vectors, an ever expanding set...
Mar 7, 2021 at 1:10 comment added IchKenneDeinenNamen Thanks for the suggestions! I may ask the question on Math SE indeed. In fact, I found this question thanks to your remark: math.stackexchange.com/questions/279445/… which may have some interesting elements. However, could you explain what you mean by "expanding kernel" in your last comment? I'm not familiar with this specific term in the context.
Mar 6, 2021 at 23:58 comment added Cosmas Zachos Actually the structure is better than bisymmetric. There is symmetry across the perpendicular and horizontal that go trough the center. That readily displays the expanding kernel.
Mar 6, 2021 at 22:33 comment added Cosmas Zachos Your A is a bisymmetric matrix. You might ask in MSE about those... They have really special properties...
Mar 6, 2021 at 22:20 comment added Cosmas Zachos ...and j+1/2 for half-integer spins. This is plausible from the bisymmetric structure...
Mar 6, 2021 at 16:58 comment added IchKenneDeinenNamen Hi @CosmasZachos, for j = 2, 0.375 is another eigenvalue and for j = 3, you also get 0.3125. As for the kernel, I can conjecture it has dimension j for integer j.
Mar 6, 2021 at 16:45 comment added Cosmas Zachos No ideas. Indeed, squaring the elements of the orthogonal matrices $d^j_{m~m'}(\pi/2)$ is the way to explore the numerics. Note for j=3/2 the eigenvalues 1 and -1/2 persist, but the kernel has increased to two null vectors... As j increases the rank of A decreases dramatically. You claim you've found different eigenvalues for j=2?
Mar 5, 2021 at 22:30 history asked IchKenneDeinenNamen CC BY-SA 4.0