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7. (a) Direct verification.

7. (a) Direct verification.

V 0 " p2lql) = (J °iy Q are orthogonal, that is, P P = \„ and Q Q = I„. Then (PQ)'(PQ) = (Q'FXPQ) = Q'(FP)Q = Q'I„Q = Q'Q = I„, so PQ is orthogonal. (d)IfP is orthogonal and c,- and Cj are two different columns of P, then C-C; is the element in row i and column j of P'P = I, so c'¡Cj = 0. If r,- and Tj are two different rows of P, then r,-iy is the element in row i and column j ofPF=I' = i, so again r,r} = 0.

ab + bd = 0; [3] ac + cd = 0; [4] be + d2 = 0. We claim that tr (A) = a+d = 0. Subtracting [4] from [1] yields a2-d2 = 0, or (a-d)(fl+d) = 0. Either a + d = 0 and we are through, or a = d. But if a = d, then [2] implies that ab = 0, so [1] implies that a3 = -abc = 0. Hence, a = 0 and tr (A) = a + d = 0 even when a = d.

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