By Boij M., Laksov D.

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Y I2 + . . 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 .

Let (X, dX ) and (Y, dY ) be metric spaces, and T a dense subset of X. Moreover, let f and g be continuous functions from X to Y . If f (x) = g(x) for all x in T , then f (x) = g(x), for all x in X. Proof: Assume that the lemma does not hold. Then there is a point x in X such that f (x) = g(x). Let ε = dY (f (x), g(x)). The balls B1 = B(f (x), 2ε ) and B2 = B(g(x), 2ε ) do not intersect, and the sets U1 = f −1 (B1 ) and U2 = g −1 (B2 ) are open in X and contain x. Since T is dense we have a point y in T contained in U1 ∩U2 .

We have that eX (t) = XeX (t). In fact, fix an elements t in K and define et : Mn (K) → Mn (K) and et : Mn (K) → Mn (K) by et (X) = eX (t) and et (X) = eX (t). We shall prove that Xet (X) = et (X). 4. 5 that it suffices to show that Xet (X) and et (X) are equal on diagonalizable matrices X. 8 that Y −1 XY et (Y −1 XY ) = Y −1 XY eY −1 XY (t) = Y −1 XeX (t)Y . 1 that et (Y −1 XY ) = eY −1 XY (t) = (Y −1 eX Y ) (t) = Y −1 eX (t)Y . Hence, 52 The exponential function to prove that et = et it suffices to prove that they are equal on diagonal matrices.

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

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