Exercises on left and right inverses; pseudoinverse � Problem 32.1: Find a right inverse for A = 1 0 1 0 1 0 � . −1 Solution: We apply the formula Aright = A T ( AA T )−1 : ⎤ 1 0 = ⎣ 0 1 ⎦ 1 0 � � 2 0 = 0 1 � � 1/2 0 = 0 1 ⎡ ⎤ 1/2 0 = ⎣ 0 1 ⎦. 1/2 0 ⎡ AT AA T ( AA T )−1 A T ( AA T )−1 ⎡ ⎤ 1/2 0 −1 Thus, Aright = ⎣ 0 1 ⎦ is one right inverse of A. We can quickly check 1/2 0 −1 that AAright = I. � 4 3 Problem 32.2: Does the matrix A = have a left inverse? A right 8 6 inverse? A pseudoinverse? If the answer to any of these questions is ”yes”, ﬁnd the appropriate inverse. � Solution: The second row of A is a multiple of the ﬁrst row, so A has rank 1 and det A = 0. Because A is a square matrix its determinant is deﬁned, and we can use the fact that det C · det D = det(CD ) to prove that A can’t have a left or right inverse. (If AB = I, then det A det B = det I implies 0 = 1.) We can ﬁnd a pseudoinverse A+ = VΣ+ U T for A. We start by ﬁnding the singular value decomposition UΣV T of A. The SVD of A was calculated in the lecture on singular value decom position, so we know that � � � � � √ � � � 2 4 3 1 .8 .6 125 0 1 = √5 . 8 6 2 −1 .6 −.8 0 0 VT Σ U A Hence, Σ+ = � √ � 1/ 125 0 and 0 0 + T A+ = V √ �Σ U �� .8 .6 1/ 125 = .6 −.8 0 �� .8 .6 1/25 0 = .6 −.8 0 0 �� .8 .6 1/25 0 = .6 −.8 0 0 � � 1 4 8 = . 125 3 6 � �� 1 1 2 √ 5 2 −1 � 1 2 2 −1 � �� 1 2 2 −1 0 0 �� To check our work, we conﬁrm that A+ reverses the operation of A on its row space using the bases we found while computing its SVD. Recall that � σj u j for j ≤ r Av j = for j > r. 0 � √ � � � 1/√5 4 1 Here u1 = and A+ u1 = √ = σ11 v1 . We can also check 125 3 2/ 5 � �� 4 8 2 1 + that A u2 = 125 = 0. 3 6 −1