Question

Please understand how the grass marked in red got three eigen values during the process.

2. Find the SVD of the following matrix .7 .6 .2 .4 H .1 .5 .9 2 .1 .3 .6 .9 2. H=UEVT 0.8685 -0.1298 -0.4793 -0.5896 -0.4272 -0.6855 -0.6508 -0.2513 U = 0.7164 0 0 1.7034 O 0.7152 0 O 0.1302 0 -0.3458 0.6849 0.4263 -0.5708 0.2191 0.0708 -0.7116 -0.6109 0.0145 -0.2198 0.3311 -0.9017 HH-I 1.os 0.63 0.79 ergen values OO こ0 0.63 1,11 .1 99 116 1.an- 2.9015), Agso,S1155, e 6.01695 EAE 01/302 0715a 1.7034 0715a O1308 A2.9015) 2a1155 oValues Ka 3

Answer 1

evaluatiny TAr-HHTL his mLana we ana the dda mirant hee 0.6 0.79 0.63 1.16 1.) 17 116 D 79 Net matiix for deteminamt -1.o5) -0 67 -079 O (A-1) -l16 (n-121) dstoinant aloy. ST ROW Eabutry tho 6-105) 6-) (>ート27)- 1-16と)-(013) Eoex (7-12)-(14X-07)」 + (0) 0x-16)-((11) +(0-63) 0-047) - D-9164 2. 33A t407-1:345 +(-079) [ +0-792) O D8763 O.7308 (07333)0-306 7) + 05)[t-2:33入+0 0641] |-1D5 (0154)o624) A 10 S-2-s8+2:5a+0-06417, +6.042) +0 0673 021 A A0 We gt 3eigen valuue 3d rde olynomial u Stnie