Linear Algebra Chapter 2. Dimension, Rank, and Linear - PowerPoint PPT Presentation

Linear Algebra Chapter 2. Dimension, Rank, and Linear Transformations Section 2.2. The Rank of a Matrix—Proofs of Theorems February 27, 2019 () Linear Algebra February 27, 2019 1 / 11

Table of contents Page 140 Number 6 1 Page 141 Number 14 2 Page 141 Number 18 3 () Linear Algebra February 27, 2019 2 / 11

Page 140 Number 6 Page 140 Number 6   0 2 3 1 − 4 4 1 4   Page 140 Number 6. Let A =  . Find (a) rank( A ), (b)   3 3 2 0  − 4 0 1 2 a basis for the row space of A , (c) a basis for the column space of A , (d) a basis for the nullspace of A . Solution. We apply the process of Note 2.2.A and row reduce A :     0 2 3 1 − 4 4 1 4 R 1 ↔ R 2 − 4 4 1 4 0 2 3 1   �   A =     3 3 2 0 3 3 2 0     − 4 0 1 2 − 4 0 1 2 () Linear Algebra February 27, 2019 3 / 11

Page 140 Number 6 Page 140 Number 6   0 2 3 1 − 4 4 1 4   Page 140 Number 6. Let A =  . Find (a) rank( A ), (b)   3 3 2 0  − 4 0 1 2 a basis for the row space of A , (c) a basis for the column space of A , (d) a basis for the nullspace of A . Solution. We apply the process of Note 2.2.A and row reduce A :     0 2 3 1 − 4 4 1 4 R 1 ↔ R 2 − 4 4 1 4 0 2 3 1   �   A =     3 3 2 0 3 3 2 0     − 4 0 1 2 − 4 0 1 2     1 − 1 − 1 / 4 − 1 1 − 1 − 1 / 4 − 1 R 1 → R 1 / ( − 4) R 3 → R 3 − 3 R 1 0 2 3 1 0 2 3 1 �   �   R 4 → R 4 + 4 R 1     3 3 2 0 0 6 11 / 4 3     − 4 0 1 2 0 − 4 0 − 2 () Linear Algebra February 27, 2019 3 / 11

Page 140 Number 6 Page 140 Number 6   0 2 3 1 − 4 4 1 4   Page 140 Number 6. Let A =  . Find (a) rank( A ), (b)   3 3 2 0  − 4 0 1 2 a basis for the row space of A , (c) a basis for the column space of A , (d) a basis for the nullspace of A . Solution. We apply the process of Note 2.2.A and row reduce A :     0 2 3 1 − 4 4 1 4 R 1 ↔ R 2 − 4 4 1 4 0 2 3 1   �   A =     3 3 2 0 3 3 2 0     − 4 0 1 2 − 4 0 1 2     1 − 1 − 1 / 4 − 1 1 − 1 − 1 / 4 − 1 R 1 → R 1 / ( − 4) R 3 → R 3 − 3 R 1 0 2 3 1 0 2 3 1 �   �   R 4 → R 4 + 4 R 1     3 3 2 0 0 6 11 / 4 3     − 4 0 1 2 0 − 4 0 − 2 () Linear Algebra February 27, 2019 3 / 11

Page 140 Number 6 Page 140 Number 6 (continued 1) Solution (continued).     1 − 1 − 1 / 4 − 1 1 − 1 − 1 / 4 − 1 R 3 → R 3 − 3 R 2 0 2 3 1 0 2 3 1 �     R 4 → R 4 + 2 R 2     0 6 11 / 4 3 0 0 − 25 / 4 0     0 − 4 0 − 2 0 0 6 0   1 − 1 − 1 / 4 − 1 R 4 → R 4 +(24 / 25) R 3 0 2 3 1 �    = H .   0 0 − 25 / 4 0  0 0 0 0 Since H is in row echelon form and has pivots in the first three columns we can apply Note 2.2.A to see that: () Linear Algebra February 27, 2019 4 / 11

Page 140 Number 6 Page 140 Number 6 (continued 1) Solution (continued).     1 − 1 − 1 / 4 − 1 1 − 1 − 1 / 4 − 1 R 3 → R 3 − 3 R 2 0 2 3 1 0 2 3 1 �     R 4 → R 4 + 2 R 2     0 6 11 / 4 3 0 0 − 25 / 4 0     0 − 4 0 − 2 0 0 6 0   1 − 1 − 1 / 4 − 1 R 4 → R 4 +(24 / 25) R 3 0 2 3 1 �    = H .   0 0 − 25 / 4 0  0 0 0 0 Since H is in row echelon form and has pivots in the first three columns we can apply Note 2.2.A to see that: (a) rank( A ) = 3 (the number of pivots in H ), () Linear Algebra February 27, 2019 4 / 11

Page 140 Number 6 Page 140 Number 6 (continued 1) Solution (continued).     1 − 1 − 1 / 4 − 1 1 − 1 − 1 / 4 − 1 R 3 → R 3 − 3 R 2 0 2 3 1 0 2 3 1 �     R 4 → R 4 + 2 R 2     0 6 11 / 4 3 0 0 − 25 / 4 0     0 − 4 0 − 2 0 0 6 0   1 − 1 − 1 / 4 − 1 R 4 → R 4 +(24 / 25) R 3 0 2 3 1 �    = H .   0 0 − 25 / 4 0  0 0 0 0 Since H is in row echelon form and has pivots in the first three columns we can apply Note 2.2.A to see that: (a) rank( A ) = 3 (the number of pivots in H ), () Linear Algebra February 27, 2019 4 / 11

Page 140 Number 6 Page 140 Number 6 (continued 2) Solution (continued). (b) a basis for the row space of A is the nonzero rows of   1 − 1 − 1 / 4 − 1 0 2 3 1   H =  ,   0 0 − 25 / 4 0  0 0 0 0 { [1 , − 1 , − 1 / 4 , − 1] , [0 , 2 , 3 , 1] , [0 , 0 , − 25 / 4 , 0] } (of course we could clean this up by multiplying the first and third vectors by 4 and getting the basis { [4 , − 4 , − 1 , − 4] , [0 , 2 , 3 , 1] , [0 , 0 , − 25 , 0] } ), () Linear Algebra February 27, 2019 5 / 11

Page 140 Number 6 Page 140 Number 6 (continued 3) Solution (continued).   0 2 3 1 − 4 4 1 4   (c) a basis for the column space of A =  is given by the   3 3 2 0  − 4 0 1 2  1 − 1 − 1 / 4 − 1  0 2 3 1   columns of A corresponding to columns of H =   0 0 − 25 / 4 0   0 0 0 0         0 2 3     − 4 4 1         which contain pivots,  ,  , .       3 3 2         4 0 1   () Linear Algebra February 27, 2019 6 / 11

Page 140 Number 6 Page 140 Number 6 (continued 4) Solution (continued). (d) For a basis for the nullspace of A , we consider x = � the homogeneous system A � 0, which has (by Theorem 1.6) the same x = � solution as H � 0. To simplify computations, we further row reduce the augmented matrix [ H | � 0]:  1 − 1 − 1 / 4 − 1 0  0 2 3 1 0 [ H | �   0] =   0 0 − 25 / 4 0 0   0 0 0 0 0  1 − 1 − 1 / 4 − 1 0  R 2 → R 2 / 2 R 1 → R 1 + R 2 0 1 3 / 2 1 / 2 0 �   � . . . R 3 → ( − 4 / 25) R 3   0 0 1 0 0   0 0 0 0 0 () Linear Algebra February 27, 2019 7 / 11

Page 140 Number 6 Page 140 Number 6 (continued 4) Solution (continued). (d) For a basis for the nullspace of A , we consider x = � the homogeneous system A � 0, which has (by Theorem 1.6) the same x = � solution as H � 0. To simplify computations, we further row reduce the augmented matrix [ H | � 0]:  1 − 1 − 1 / 4 − 1 0  0 2 3 1 0 [ H | �   0] =   0 0 − 25 / 4 0 0   0 0 0 0 0  1 − 1 − 1 / 4 − 1 0  R 2 → R 2 / 2 R 1 → R 1 + R 2 0 1 3 / 2 1 / 2 0 �   � . . . R 3 → ( − 4 / 25) R 3   0 0 1 0 0   0 0 0 0 0 () Linear Algebra February 27, 2019 7 / 11

Page 140 Number 6 Page 140 Number 6 (continued 5) Solution (continued).  1 0 5 / 4 − 1 / 2 0  R 1 → R 1 + R 2 0 1 3 / 2 1 / 2 0 �   . . .   0 0 1 0 0   0 0 0 0 0  1 0 0 − 1 / 2 0  R 1 → R 1 − (5 / 4) R 3 0 1 0 1 / 2 0 �    . R 2 → R 2 − (3 / 2) R 3   0 0 1 0 0  0 0 0 0 0 Returning to a system of equations, . . . () Linear Algebra February 27, 2019 8 / 11

Page 140 Number 6 Page 140 Number 6 (continued 5) Solution (continued).  1 0 5 / 4 − 1 / 2 0  R 1 → R 1 + R 2 0 1 3 / 2 1 / 2 0 �   . . .   0 0 1 0 0   0 0 0 0 0  1 0 0 − 1 / 2 0  R 1 → R 1 − (5 / 4) R 3 0 1 0 1 / 2 0 �    . R 2 → R 2 − (3 / 2) R 3   0 0 1 0 0  0 0 0 0 0 Returning to a system of equations, . . . () Linear Algebra February 27, 2019 8 / 11

Page 140 Number 6 Page 140 Number 6 (continued 6) Solution (continued). . . . − (1 / 2) x 4 = 0 or x 1 = (1 / 2) x 4 x 1 x 2 +(1 / 2) x 4 = 0 x 2 = − (1 / 2) x 4 = 0 x 3 = 0 x 3 0 = 0 x 4 = x 4 . x 1 = (1 / 2)(2 r ) = r = ( − 1 / 2)(2 r ) = − r x 2 With r = x 4 / 2 as a free variable we have . So x 3 = 0 x 4 = 2 r  �   1  �    �  − 1   x = �   � r ∈ R the general solution set for the system A � 0 is and r   � 0   �    �  2   � so a basis for the nullspace of A is { [1 , − 1 , 0 , 2] T } . � () Linear Algebra February 27, 2019 9 / 11

Page 140 Number 6 Page 140 Number 6 (continued 6) Solution (continued). . . . − (1 / 2) x 4 = 0 or x 1 = (1 / 2) x 4 x 1 x 2 +(1 / 2) x 4 = 0 x 2 = − (1 / 2) x 4 = 0 x 3 = 0 x 3 0 = 0 x 4 = x 4 . x 1 = (1 / 2)(2 r ) = r = ( − 1 / 2)(2 r ) = − r x 2 With r = x 4 / 2 as a free variable we have . So x 3 = 0 x 4 = 2 r  �   1  �    �  − 1   x = �   � r ∈ R the general solution set for the system A � 0 is and r   � 0   �    �  2   � so a basis for the nullspace of A is { [1 , − 1 , 0 , 2] T } . � () Linear Algebra February 27, 2019 9 / 11