Exceptional Lie Algebras by N. Jacobson
By N. Jacobson
This e-book is the single position to discover plenty of theorems on unheard of Lie algebras (and consequently additionally on remarkable algebraic groups).
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Turning to the equivalence between (i) and (ii), let us for notational convenience put A = Ai ∩ (Ai ∩ Aj )⊥ , B = A⊥ j + (Ai ∩ Aj ), C = Ai ∩ A⊥ j . We are going to show that A = C. If (ii) is true, A ⊆ B. 3 (i) implies B = A + A⊥ ∩ B, and we always have that B = C + C⊥ ∩ B. However, C⊥ ∩ B = (A⊥ + B⊥ ) ∩ B = A⊥ ∩ B, giving us A = C. Thus (ii) implies (i). The converse is trivial. 1 (v) that if (iii) holds, ⊥ Ai ∩ (Ai ∩ Aj )⊥ = Ai ∩ (Ai ∩ Aj )⊥ ∩ A⊥ j = Ai ∩ Aj . The converse is obvious. 8 (ii) expresses orthogonality of Ai and Aj modulo Ai ∩ Aj .
15) A− = A− AA− . 6. A g-inverse A− : n × m is a reﬂexive g-inverse if and only if r(A− ) = r(A). Proof: By deﬁnition of a reﬂexive g-inverse, r(A− ) = r(A− A) as well as r(A) = r(AA− ). 4 (v) follows that r(AA− ) = tr(AA− ) = tr(A− A) = r(A− A) and thus reﬂexivity implies r(A) = r(A− ). 3 (x). Then r(A− − A− AA− ) = r(A− ) + r(Im − AA− ) − m = r(A− ) − r(AA− ) = r(A− ) − r(A) = 0 which establishes the theorem. Note that for a general g-inverse r(A) ≤ r(A− ), whereas reﬂexivity implies equality of the ranks.
Belong to V, where the operations ” + ” (sum of vectors) and ” · ” (multiplication by scalar) are deﬁned so that x + y ∈ V, αx = α · x ∈ V, where α belongs to some ﬁeld K and x + y = y + x, (x + y) + z = x + (y + z), there exists a unique null vector 0 in the space so that, for all x ∈ V, x + 0 = x, for every x ∈ V there exists a unique −x ∈ V so that x + (−x) = 0, 1 · x = x, α(βx) = (αβ)x, α, β ∈ K, (α + β)x = αx + βx, α(x + y) = αx + αy. If these conditions are satisﬁed we say that we have a vector space V over the ﬁeld K.