By Boij M., Laksov D.

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**Extra resources for An introduction to algebra and geometry via matrix groups**

**Example text**

4. Let X be a matrix in Mn (C). Then there exists a matrix Y , complex numbers di and ei , for i = 1, . . , n, with the ei all different, and a real positive number ε such that, for all nonzero t ∈ C with |t| < ε, we have that X + tY is diagonalizable and with diagonal matrix whose (i, i)’th coordinate is di + tei , for i = 1, . . , n. Proof: The proposition clearly holds when n = 1. We shall proceed by induction on n. Assume that the proposition holds for n − 1. Choose an eigenvalue d1 of X and a nonzero eigenvector x1 for d1 .

Consequently (t exp X)S exp X = S, and exp X is in GS (K), as asserted. To prove the second assertion about GS (K) we take A in GS (K). Then t ASA = S or equivalently, S −1t ASA = In . Consequently we have that log((S −1t AS)A) = 0. We have that S −1t AS is the inverse matrix of A, and hence that S −1t AS and A commute. Since the map of Gln (K) that sends A to t A is continuous, as is the map that sends A to S −1 AS, we can choose U such that log is defined on S −1 (t A)S for all A in U . 6 that log((S −1t AS)A) = log(S −1t AS)+log A = S −1 (log t A)S +log A = S −1 (t log A)S + log A.

4 that we have dimK W ⊥ = n−1. For y ∈ W ⊥ we have that sx (y) = y and we have that sx (x) = −x. In particular s2x is the identity map. 1) x, z − 2 y, x = y, z − 2 x, x x, x y, x z, x +4 x, x = y, z . x, x 2 Since det sx = −1, we have that sx ∈ O(V ) \ SO(V ). 2). 3. Let x and y be two elements of V such that x, x = y, y = 0. Then there is a linear map V → V , which takes x to y and which is a product of at most 2 reflections of the form sz . Proof: Assume that x, y = x, x = y, y . Then x − y, x − y = 2( x, x − y ) = x,x−y 2( x, x − x, y ) = 0.