Question

Q 4) Linear Algebra. Please show me the work clearly.

.. 8 6.5 ] | 8 6.5 A = -4 -9.5 -9.5] A singular value decomposition of A is [ 0.5 0.5 -0.5 -0.57 20 0] 0.5 0.5 0.5 0.5 || 0 5 | 0.6 0.8 ] -0.5 0.5 -0.5 0.5 || 0 0 (0.8 -0.6] (-0.5 0.5 0.5 -0.5][ 0 0] Find bases for the column space of A and for the null space for AT. (help with entering vector) Col(A) = span{ Null(AT) = span . Given the vector A = b= Find the projection of the vector b on to the column space of A: Projcola (b) = Find the projection of the vector b on to the orthogonal complement of column space of A:

Answer 1

S.9 6-5 -9.5 |-4 -4 -9.5 -4 AT- 6.5 -g.5 -9.5 6.5 Null spacl of AT -4 8. -4 19 - 19 13 2. t (-1, 1,0, 0) +s (0, 0, -1, 1) basis of Null spacu of AT codeemn spacl of A 8 6-S 8 6.5 8 6.5 -4 -9.5 -4 -95. 8 65 O -25/4 -15/4 o -25/4 -25/4

8 13/2 o 25/4 hene co lumnspoce is iven by the equaliou A, B GR -19142 basi's {v, ,V. } 13/2 -19/2 -19/2 Find the pro jection of the vectr on the coluon spacen where be (4, -1, 2, -1)T Proj v, b <V, 6) <, Vi> 32-9 - S +4 (४, ४,-५- (8,8, -4,-4)' ИПо %3D 4 Sio |ч9ro -2vo -20/ . -2Vo Proj v, b (26 - 132 -19 -19/2) (13/2 1/2-1% V, b V2 = Zv, V 169+169 301 1 381 -1 (13/2 ,13/2 -19 -19/2)' 15 F17/30 -17/30 (1/30