314 CHAPTER 12. INNER PRODUCT SPACES, LEAST SQUARES

Corollary 12.3.2 Let A ∈L (X ,Y ) where X and Y are two inner product spaces of finitedimension or else Hilbert spaces. Then there exists a unique A∗ ∈L (Y,X) , the boundedlinear transformations, such that

(Ax,y)Y = (x,A∗y)X (12.5)

for all x ∈ X and y ∈ Y. The following formula holds

(αA+βB)∗ = αA∗+βB∗

Also, (A∗)∗ = A.

Proof: Let fy ∈L (X ,F) be defined as

fy (x)≡ (Ax,y)Y .

This is linear and ∣∣ fy (x)∣∣= |(Ax,y)Y | ≤ |Ax| |y| ≤ (||A|| |y|) |x|

Then by the Riesz representation theorem, there exists a unique element of X , A∗ (y) suchthat

(Ax,y)Y = (x,A∗ (y))X .

It only remains to verify that A∗ is linear. Let a and b be scalars. Then for all x ∈ X ,

(x,A∗ (ay1 +by2))X ≡ (Ax,(ay1 +by2))Y

≡ a(Ax,y1)+b(Ax,y2)≡a(x,A∗ (y1))+b(x,A∗ (y2)) = (x,aA∗ (y1)+bA∗ (y2)) .

Since this holds for every x, it follows

A∗ (ay1 +by2) = aA∗ (y1)+bA∗ (y2)

which shows A∗ is linear as claimed.Consider the last assertion that ∗ is conjugate linear.(

x,(αA+βB)∗ y)≡ ((αA+βB)x,y)

= α (Ax,y)+β (Bx,y) = α (x,A∗y)+β (x,B∗y)

= (x,αA∗y)+(

x,βA∗y)=(

x,(

αA∗+βA∗)

y).

Since x is arbitrary,(αA+βB)∗ y =

(αA∗+βA∗

)y

and since this is true for all y,

(αA+βB)∗ = αA∗+βA∗.

Finally, (A∗x,y) = (y,A∗x) = (Ay,x) = (x,Ay) while (A∗x,y) =(x,(A∗)∗ y

)and so for

all x, (x,(A∗)∗ y−Ay

)= 0

and so (A∗)∗ = A. ■

314 CHAPTER 12. INNER PRODUCT SPACES, LEAST SQUARESCorollary 12.3.2 Let A € & (X,Y) where X and Y are two inner product spaces of finitedimension or else Hilbert spaces. Then there exists a unique A* € & (Y,X), the boundedlinear transformations, such that(Ax,y)y = (x,A*Y)x (12.5)for allx € X and y € Y. The following formula holds(a@A + BB)* = @A* + BB*Also, (A*)* =A.Proof: Let f, € 2 (X,IF) be defined asfy (x) = (Axy)y-This is linear and| fy (x) | = [(Ax.y)y| < JAxl [yl < ({IATL Lvl) LaThen by the Riesz representation theorem, there exists a unique element of X, A* (y) suchthat(Ax,y)y = (*,A* (y))x-It only remains to verify that A* is linear. Let a and b be scalars. Then for all x € X,(x,A* (ay1 + by2))x = (Ax, (ay + by2))y= a(Ax,y1) +b (Ax,y2) =a(x,A* (y1)) +b (x,A* (y2)) = («,aA* (y1) + BA* (y2)).Since this holds for every x, it followsA* (ay; + by2) = aA* (vy) + DA* (y2)which shows A’ is linear as claimed.Consider the last assertion that * is conjugate linear.(x, (A + BB)*y) = ((@A + BB)x,y)= a(Ax,y) +B (Bx,y) = a (x,A*y) + B (x, B’y)(x, @A*y) + («,Ba‘y) = (s. (aa +BA") y) .Since x is arbitrary,(aA + BB)" y= (GA* + BA") yand since this is true for all y,(@A+ BB)* = G@A* + BA*.Finally, (A*x,y) = (y,A*x) = (Ay,x) = (x,Ay) while (A*x,y) = (x, (A*)* y) and so forall x,(x, (A*)*y—Ay) =0and so (A*)* =A.