318 CHAPTER 12. SELF ADJOINT OPERATORS

12.12 The Moore Penrose Inverse

The particular solution of the least squares problem given in 12.18 is important enough thatit motivates the following definition.

Definition 12.12.1 Let A be an m × n matrix. Then the Moore Penrose inverse of A,denoted by A+ is defined as

A+ ≡ V

(σ−1 0

0 0

)U∗.

Here

U∗AV =

(σ 0

0 0

)as above.

Thus A+y is a solution to the minimization problem to find x which minimizes |Ax− y| .In fact, one can say more about this. In the following picture My denotes the set of leastsquares solutions x such that A∗Ax = A∗y.

Myx

A+(y)

ker(A∗A)

Then A+ (y) is as given in the picture.

Proposition 12.12.2 A+y is the solution to the problem of minimizing |Ax− y| for all xwhich has smallest norm. Thus∣∣AA+y − y

∣∣ ≤ |Ax− y| for all x

and if x1 satisfies |Ax1 − y| ≤ |Ax− y| for all x, then |A+y| ≤ |x1| .

Proof: Consider x satisfying 12.17, equivalently A∗Ax =A∗y,(σ2 0

0 0

)V ∗x =

(σ 0

0 0

)U∗y

which has smallest norm. This is equivalent to making |V ∗x| as small as possible becauseV ∗ is unitary and so it preserves norms. For z a vector, denote by (z)k the vector in Fk

which consists of the first k entries of z. Then if x is a solution to 12.17(σ2 (V ∗x)k

0

)=

(σ (U∗y)k

0

)

318 CHAPTER 12. SELF ADJOINT OPERATORS12.12 ‘The Moore Penrose InverseThe particular solution of the least squares problem given in 12.18 is important enough thatit motivates the following definition.Definition 12.12.1 Let A be an m xn matrix. Then the Moore Penrose inverse of A,denoted by At is defined asa! 0AtT=V U*.0 Ouxav= (7 °0 0Thus Aty is a solution to the minimization problem to find x which minimizes |Ax — y|.In fact, one can say more about this. In the following picture My denotes the set of leastsquares solutions x such that A* Ax = A*y.Hereas above.Then At (y) is as given in the picture.Proposition 12.12.2 Aty is the solution to the problem of minimizing |Ax — y| for all xwhich has smallest norm. Thus|AATy — y| < |Ax—y| for all xand if x, satisfies |Ax; — y| < |Ax —y| for all x, then |ATy| < |xi|.Proof: Consider x satisfying 12.17, equivalently A* Ax =A’*y,2o- 0 Vexal 7 0 U*y0 0 0 0which has smallest norm. This is equivalent to making |V*x| as small as possible becauseV* is unitary and so it preserves norms. For z a vector, denote by (z), the vector in F*which consists of the first & entries of z. Then if x is a solution to 12.17( 0? (V*x), _ ( o(U*Y),0 0