Question: this is wha ti got 1.Compability: a (Bu) = (ap)u For all a, B E F and u E V a(Bu) = (ap)u Let u

this is wha ti got

1.Compability: a (Bu) = (ap)u For all a, B E F and u E V a(Bu) = (ap)u Let u = a + bv2 , with a, be Q a, BEQ Let's check both sides: Left- hand - side a(Bu) = a[B(a + bv2)] = a [Ba + Bbv2 ] = aBa + aBbv2 Right -hand side (aB)u = (aB) (a + bv2) = aBa + aBbv2 Bothe sides are equal, then compatibility axioms holds

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