An introduction to algebra and geometry via matrix groups by Boij M., Laksov D.

By Boij M., Laksov D.

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Then 1 = x , y = x, y = x , y . Consequently it suffices to find two more transvections whose composition maps y to y and that fix x . If y , y = 0, we let Ψ (z) = z + a y − y , z (y − y ) with a = y , y . Then we have that Ψ (y ) = y , and Ψ (x ) = x , because y − y , x = 1 − 1 = 0. On the other hand, when y , y = 0, we The center of the matrix groups 29 have that 1 = x , y = x , x + y and y , x + y = 0 = y , x + y , so we can first find a composition of two transvections that map the pair (x , y ) to (x , x + y ) and then find a composition of two transvections that map the latter pair to (x , y ).

2. Let (X1 , d1 ), . . , (Xm , dm ) be metric spaces. (a) Show that the Cartesian product X = X1 × · · · × Xm with the function d : X × X → R defined by d((x1 , . . , xm ), (y1 , . . , ym )) = d1 (x1 , y1 ) + · · · + dm (xm , ym ), is a metric space. (b) When X1 = · · · = Xm and di , for i = 1, . . 2, the metric is called the Hamming metric on X. Show that d((x1 , . . , xm ), (y1 , . . , ym )) is the number of indices i such that xi = yi . (c) When X1 = · · · = Xm = K, where K is the real or complex numbers, we have that X = VKm , and we call the metric, the taxi metric.

0 .. . log(an ) (v) log(A) log(B) = log(B) log(A), when AB = BA, and log(A), log(B) and log(AB) are defined. 8. For the last assertion we note that when AB = BA the partial sums logm (A) logm (B) and logm (B) logm (A) are actually equal. 13. The logarithm defines a continuous map log : B(In , 1) → Mn (K). 3). 4 we shall show that the 2 logarithmic function is analytic. Hence, in particular, it is differentiable with an analytic derivative. 1. Determine the matrices exp ( 10 11 ), exp ( 14 11 ), and exp 1 1 −1 3 .

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