N y a = -12, 6= 4,c= -25 a+b=b+a
Answers
Answer:
Let
U=12[0−110],V=[0110]
Then
UV=12[100−1],VU=12[−1001]
hence
UV−VU=[100−1]
and its square is the identity.
To create a 2n×2n matrix with the same characteristic, let An consist of n copies of U along the diagonal in blocks, and let Bn be n copies of V along the diagonal in blocks. Then
(AnBn−BnAn)2=I2n.
This leaves open the question for matrices of odd size. Offhand, I'm guessing that there might be no solution in that case, but perhaps there's something similar to this one, where AB−BA turns out to be rotation by π about the axis (1,1,1) or something.
Post-comment remarks Will Jagy closes this out by noting that if n is odd, and you have a solution consisting of matrices A and B, then you can let
W=AB−BA
This can be put in Jordan normal form so that
P−1WP
is a Jordan-block matrix. And its square is the identity, because it's P−1W2P, and W2 is assumed to be I. The square of a jordan block is the identity only if the block itself is diagonal, hence P−1WP is diagonal.
Replacing A and B with A′=P−1AP and B′=P−1BP, we have
A′B′−B′A′=P−1WP
is diagonal, and its square is still the identity. Hence all its diagonal entries are ±1, so its trace is ±3 or ±1. But in general, tr(AB−BA)=0, so we have a contradiction.
Conclusion: for odd n, there are no such matrices A and B.
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Answer:
the answer is
a+b=b+a=0