20_hwc1SolnsODDA

20_hwc1SolnsODDA - AB = BA + [ A, B ]. (a) Compute [ A, B ]...

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(b) Find all possible values of a b and c so that A 2 = B where B = h 1 3 0 4 i . How many different sets of values for a b and c are there for which A 2 = B ? (The matrices A that you compute here are square roots of the matrix B .) SOLUTION (a) ± a 2 ab + bc 0 c 2 ² . (b) We see by comparing the upper left entries that we have to have a 2 = 1, so we have to have a = ± 1. We see by comparing the lower right entries that we have to have c 2 = 4, so we have to have c = ± 2. This gives us a total of four choices for the values of a and c . But then, whichever of these four we choose, we can go on to choose exactly one value of b so that we get equality in the upper right entries. (The lower left entries are always zero, and so they take care of themselves.) Thus, there are four different “sqaure roots” of the matrix B : ± 1 1 0 2 ² ± 1 - 3 0 - 2 ² ± - 1 1 0 2 ² and ± - 1 - 1 0 - 2 ² . 3.9 Given two n × n matrices A and B , the commutator of A and B is defined to be the n × n matrix AB - BA , and is denoted by [ A, B ]. The reason for this is that AB = BA if and only if [ A, B ] = 0. More generally, but also directly from the definition,
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Unformatted text preview: AB = BA + [ A, B ]. (a) Compute [ A, B ] for A = h 1 1 i and B = h 1 2 i . (b) For the same matrices A and B , compute [ A, [ A, B ]]. (c) Using only the associative property of matrix multiplication, show that for any three n n matrices A , B and C , [ A, [ B, C ]] + [ B, [ C, A ]] + [ C, [ A, B ]] = 0 . This is known as Jacobis identity . (We wont make use of it later. For our purposes, checking it is simply a good exercise in working with matrices.) SOLUTION (a) [ A, B ] = 1-1 . (b) Let C = [ A, B ]. From part (a) you know that C = 1-1 Now you just compute the commutator the way you did previously: [ A, [ A, B ]] = AC-CA = -2 2 For (a) , we work out, using the denitions that we just practiced with: [ A, [ B, C ]] = ABC + CBA-ACB-BCA 21 /september/ 2005; 22:09 21...
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