By Boij M., Laksov D.

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**Example text**

Consequently, we have x y y sin y that exp(yX) = −cos sin y cos y . 12). Then we have that Φ(exp(iy)) = exp(Φ(iy)). In the last formula the exponential function on the left is the usual exponential function for complex numbers and the one to the right the exponential function for matrices. 6. The exponential function defines a continuous map exp : Mn (K) → Mn (K). Indeed, we have seen that expm (X) ≤ exp( X ). Let B(Z, r) be a ball in Mn (K), and choose Y in Mn (K) such that Z + r ≤ Y . Then, for any X in B(Z, r), we have that X ≤ X − Z + Z ≤ r + Z ≤ Y .

1, is complete. Proof: Let xi = (ai1 , . . , ain ) be a Cauchy sequence in VKn . Given ε there is an integer m such that xi − xj = maxk |aik − ajk | < ε, when i, j > m. Consequently, the sequences a1k , a2k , . . are Cauchy in K, for k = 1, . . , n. Since the real and complex numbers are complete we have that these sequences converge to elements a1 , . . , an . It is clear that x1 , x2 , . . converges to x = (a1 , . . , an ). 3. For X in Mn (K) and m = 0, 1, . . , let expm (X) be the matrix 1 1 1 m expm (X) = In + X + X 2 + · · · + X .

8) is of the form S= 0 Jm , −Jm 0 where Jm be the matrix in Mm (K) with 1 on the antidiagonal, that is the elements aij with i + j = m + 1 are 1, and the remaining coordinates 0. Moreover the basis can be chosen so that it contains any given non-zero vector x. Proof: If n = 1 there is no non-degenerate alternating form. So assume that n > 1. Let e1 be an arbitrary non-zero vector. Since the form is non-degenerate there is a vector v such that e1 , v = 0. Let en = e11,v v. Then e1 , en = 1. Let W = Ke1 + Ken be the subspace of V spanned by e1 and en .