Transcription of SVD computation example A U V T A 3 2 2 AA
1 SVD computation exampleExample:Find the SVD ofA,U VT, whereA=(3 2 22 3 2).First we compute the singular values iby finding the eigenvalues (17 88 17).The characteristic polynomial isdet(AAT I) = 2 34 + 225 = ( 25)( 9), sothe singular values are 1= 25 = 5 and 2= 9 = we find the right singular vectors (the columns ofV) by finding an orthonormalset of eigenvectors ofATA. It is also possible to proceed by finding the left singularvectors (columns ofU) instead. The eigenvalues ofATAare 25, 9, and 0, and sinceATAis symmetric we know that the eigenvectors will be = 25, we haveATA 25I= 12 12 212 12 22 2 17 which row-reduces to 1 1 00 0 10 0 0.
2 A unit-length vector in the kernel of that matrixisv1= 1/ 21/ 20 .For = 9 we haveATA 9I= 4 12 212 4 22 2 1 which row-reduces to 1 0 140 1140 0 0 .A unit-length vector in the kernel isv2= 1/ 18 1/ 184/ 18 .For the last eigenvector, we could compute the kernel ofATAor find a unit vectorperpendicular tov1andv2. To be perpendicular tov1= abc we need a= the condition thatvT2v3= 0 becomes 2a/ 18 + 4c/ 18 = 0 or a= 2c. Sov3= a a a/2 and for it to be unit-length we needa= 2/3 sov3= 2/3 2/3 1/3 .12So at this point we know thatA=U VT=U(5 0 00 3 0) 1/ 21/ 201/ 18 1/ 18 4/ 182/3 2/3 1/3 .Finally, we can computeUby the formula ui=Avi, orui=1 Avi.
3 This givesU=(1/ 2 1/ 21/ 2 1/ 2). So in its full glory the SVD is:A=U VT=(1/ 2 1/ 21/ 2 1/ 2)(5 0 00 3 0) 1/ 21/ 201/ 18 1/ 18 4/ 182/3 2/3 1/3.
