Multivariable Advanced Calculus

3.8. INNER PRODUCT AND NORMED LINEAR SPACES 61Consider the second claim. From the first claim,(Lx · y) = (x·L ∗ y) = ( (L ∗ ) ∗ x · y )and since this holds for all y, it follows Lx = (L ∗ ) ∗ x.Consider the third claim.(x· (aL + bM) ∗ y ) = ((aL + bM) x · y) = a (Lx · y) + b (Mx · y)and (x·(aL ∗ + bM ∗) y ) = a (x·L ∗ y) + b (x·M ∗ y) = a (Lx · y) + b (Mx · y)and since ( x· (aL + bM) ∗ y ) = ( x· (aL ∗ + bM ∗) y ) for all x, it must be that(aL + bM) ∗ y = ( aL ∗ + bM ∗) yfor all y which yields the third claim.Consider the fourth.(x· (ML) ∗ y ) = ((ML) x · y) = (Lx·M ∗ y) = (x·L ∗ M ∗ y)Since this holds for all x, y the conclusion follows as above. This proves the theorem.Here is a very important example.Example 3.8.17 Suppose F ∈ L (H 1 , H 2 ) . Then F F ∗ ∈ L (H 2 , H 2 ) and is self adjoint.To see this is so, note it is the composition of linear transformations and is thereforelinear as stated. To see it is self adjoint, Proposition 3.8.16 implies(F F ∗ ) ∗ = (F ∗ ) ∗ F ∗ = F F ∗In the case where A ∈ L (F n , F m ) , considering the matrix of A with respect to theusual bases, there is no loss of generality in considering A to be an m × n matrix,(Ax) i= ∑ jA ij x j .Then in terms of components of the matrix, A,(A ∗ ) ij= A ji .You should verify this is so from the definition of the usual inner product on F k . Thefollowing little proposition is useful.Proposition 3.8.18 Suppose A is an m × n matrix where m ≤ n. Also supposedet (AA ∗ ) ≠ 0.Then A has m linearly independent rows and m independent columns.Proof: Since det (AA ∗ ) ≠ 0, it follows the m × m matrix AA ∗ has m independentrows. If this is not true of A, then there exists x a 1 × m matrix such thatHencexA = 0.xAA ∗ = 0and this contradicts the independence of the rows of AA ∗ . Thus the row rank of Aequals m and by Corollary 3.5.20 this implies the column rank of A also equals m. Thisproves the proposition.