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T    A (e ) eTi Aej = eTi  = e i  Aij  = Aij  ik j k     .  ..    ..  . Amk (ej )k Amj by the first part of the lemma. This proves the lemma. 4 Let L : Fn → Fm be a linear transformation. Then there exists a unique m × n matrix, A such that Ax = Lx for all x ∈ Fn . 4. SUBSPACES AND SPANS 53 Proof: By the lemma, (Lx)i = eTi Lx = eTi xk Lek = eTi Lek xk . Let Aik = eTi Lek , to prove the existence part of the theorem. To verify uniqueness, suppose Bx = Ax = Lx for all x ∈ Fn .

2722 × 10−5 in radians per second. 2722 × 10−5 2 |rB | . Clearly this is not worth considering in the presence of the acceleration due to gravity which is approximately 32 feet per second squared near the surface of the earth. If the acceleration a, is due to gravity, then aB = a − Ω× (Ω × R) − 2Ω × vB = ≡g − Note that GM (R + rB ) |R + rB | 3 − Ω× (Ω × R) − 2Ω × vB ≡ g − 2Ω × vB . 2 Ω× (Ω × R) = (Ω · R) Ω− |Ω| R and so g, the acceleration relative to the moving coordinate system on the earth is not directed exactly toward the center of the earth except at the poles and at the equator, 64 MATRICES AND LINEAR TRANSFORMATIONS although the components of acceleration which are in other directions are very small when compared with the acceleration due to the force of gravity and are often neglected.

This means that whenever something satisfies these properties, the Cauchy Schwartz inequality holds. There are many other instances of these properties besides vectors in Fn . The Cauchy Schwartz inequality allows a proof of the triangle inequality for distances in Fn in much the same way as the triangle inequality for the absolute value. 25) and equality holds if and only if one of the vectors is a nonnegative scalar multiple of the other. 26) Proof : By properties of the dot product and the Cauchy Schwartz inequality, 2 |a + b| = (a + b) · (a + b) = (a · a) + (a · b) + (b · a) + (b · b) 2 = |a| + 2 Re (a · b) + |b| 2 2 2 2 ≤ |a| + 2 |a · b| + |b| ≤ |a| + 2 |a| |b| + |b| 2 = (|a| + |b|) .

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