Today also there is another linear algebra question given below.?

By hypothesis, A = PJP^(-1) for some invertible P and J being the Jordan Normal Form of A. Moreover, we have that |A| = |J| = 8, and tr(A) = tr(J) = 2?2. Since A^2 = 2I, we have (PJP^(-1))^2 = 2I P J^2 P^(-1) = 2I Multiplying on the left by P^(-1) and on the right by P yields J^2 = P^(-1) (2I) P J^2 = 2I, since 2I commutes with P and P^(-1).

---------- This implies that J is diagonal, because the square of a Jordan block (with 1's on the superdiagonal) is not a diagonal matrix unless the diagonal is 0 (this possibility is rule out since |A| = |J| = 8 is nonzero). Since J^2 = 2I, and J is diagonal, this implies that J is a diagonal matrix whose diagonal entries are ±?2, and possibly a mix of the two possibilities. Since |J|= 8 = product of diagonal entries of J, and tr(J) = 2?2 = sum of diagonal entries of J: J has an even number of -?2's on the diagonal.

Let a = number of -?2's and be = number of?2's on the diagonal of J. So, we have (i) n = a+b (ii) |J| = 8 = (-?2)^a * (?2)^b =?2^n (since a is even and a + be = n) (iii) tr J = 2?2 = a * -?2 + be *?2 ==> 2 = -a + b.

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