gibson:teaching:fall-2014:iam961:svddemo
IAM 961 SVD demo
A numerical demo of how uniqueness works for the SVD of real-valued matrices, conducted in Matlab.
% Form "random" matrix A with known SVD: A = U1 S1 V1'
% Use QR decomposition on random matrix to et random unitary matrix
% The QR decomposition of matrix A is a factorization A = QR with Q
% unitary Q and upper-triangular R. The QR decomp of a matrix with
% normally distributed random elements with produce a unitary matrix
% with its orthogonal columns in an isotropically random orientation.
[U1,tmp] = qr(randn(5,5));
[V1,tmp] = qr(randn(5,5));
>> U1
U1 =
-0.1735 -0.2978 0.4339 0.8304 0.0575
-0.5919 -0.1608 -0.7489 0.2185 -0.1230
0.7291 0.1574 -0.4449 0.4549 -0.1970
-0.2783 0.7273 0.1992 0.1386 -0.5785
-0.1029 0.5760 -0.1147 0.1910 0.7798
% Demonstrate that U1 is unitary, in a variety of ways
>> U1 * U1'
ans =
1.0000 -0.0000 0.0000 -0.0000 -0.0000
-0.0000 1.0000 -0.0000 -0.0000 -0.0000
0.0000 -0.0000 1.0000 0.0000 0.0000
-0.0000 -0.0000 0.0000 1.0000 0.0000
-0.0000 -0.0000 0.0000 0.0000 1.0000
>> U1 * U1' - eye(5)
ans =
1.0e-15 *
0 -0.0278 0.0798 -0.0347 -0.0069
-0.0278 0 -0.0798 -0.2914 -0.1249
0.0798 -0.0798 -0.1110 0.0833 0.0278
-0.0347 -0.2914 0.0833 0.4441 0.1665
-0.0069 -0.1249 0.0278 0.1665 0.2220
>> norm(U1 * U1' - eye(5))
ans =
6.9598e-16
% Show V1 is unitary
>> norm(V1 * V1' - eye(5))
ans =
3.0661e-16
% Now construct S1 with known diagonal entries
>> S1 = diag([127, 34, 3.5, 0.72, 0.00573])
S1 =
127.0000 0 0 0 0
0 34.0000 0 0 0
0 0 3.5000 0 0
0 0 0 0.7200 0
0 0 0 0 0.0057
>> A = U1*S1*V1'
A =
15.3394 15.7662 -2.5713 9.0749 -4.2474
50.8731 39.5276 -13.2333 13.7956 -34.2276
-61.9380 -46.1719 16.7968 -20.9126 43.7778
22.1138 1.4313 -12.9916 -1.3913 -34.6287
7.6613 -5.5310 -7.2904 -4.8075 -19.7186
% Notation:
% use U1,S1,V1 for the constructed, known SVD
% use U2,S2,V2 for the calculated SVD
% Now calculate the SVD of A
>> [U2,S2,V2] = svd(A);
% Verify that the calculated SVD actually equals A
>> U2 * S2 * V2'
ans =
15.3394 15.7662 -2.5713 9.0749 -4.2474
50.8731 39.5276 -13.2333 13.7956 -34.2276
-61.9380 -46.1719 16.7968 -20.9126 43.7778
22.1138 1.4313 -12.9916 -1.3913 -34.6287
7.6613 -5.5310 -7.2904 -4.8075 -19.7186
>> A
A =
15.3394 15.7662 -2.5713 9.0749 -4.2474
50.8731 39.5276 -13.2333 13.7956 -34.2276
-61.9380 -46.1719 16.7968 -20.9126 43.7778
22.1138 1.4313 -12.9916 -1.3913 -34.6287
7.6613 -5.5310 -7.2904 -4.8075 -19.7186
>> norm(A-U2*S2*V2')
ans =
7.6825e-14
>> norm(A-U1*S1*V1')
ans =
0
% Now compare the factors of the known and calculated SVDs
>> U1
U1 =
-0.1735 -0.2978 0.4339 0.8304 0.0575
-0.5919 -0.1608 -0.7489 0.2185 -0.1230
0.7291 0.1574 -0.4449 0.4549 -0.1970
-0.2783 0.7273 0.1992 0.1386 -0.5785
-0.1029 0.5760 -0.1147 0.1910 0.7798
>> U2
U2 =
0.1735 -0.2978 0.4339 0.8304 0.0575
0.5919 -0.1608 -0.7489 0.2185 -0.1230
-0.7291 0.1574 -0.4449 0.4549 -0.1970
0.2783 0.7273 0.1992 0.1386 -0.5785
0.1029 0.5760 -0.1147 0.1910 0.7798
>> U1-U2
ans =
1.0e-15 *
-0.1110 0.4441 -0.2776 0 0
0.1110 -0.0833 0.1110 -0.1110 0.0971
-0.1110 0.2498 0 0 -0.0555
-0.0555 -0.2220 -0.1388 -0.3886 -0.1110
-0.0278 0 0.0139 -0.1110 0
% Amazing: U1 and U2 are actually equal, even though the SVD
% is unique only up to +/- 1 in the columns of U and V.
% Will demonstrate this by changing the sign of 1st columns of U2
% and V2 and showing that the modified SVD still satisfies A = U2 S2 V2'
>> U2(:,1)= -1*U2(:,1);
>> V2(:,1)= -1*V2(:,1);
>> V1
V1 =
-0.6683 -0.0590 -0.1033 0.2860 0.6763
-0.4695 -0.6019 -0.3716 -0.1836 -0.4955
0.1960 -0.2385 -0.1227 -0.8035 0.4939
-0.1898 -0.3528 0.9098 -0.1026 -0.0359
0.5085 -0.6730 -0.0920 0.4778 0.2277
>> V2
V2 =
0.6683 -0.0590 -0.1033 0.2860 0.6763
0.4695 -0.6019 -0.3716 -0.1836 -0.4955
-0.1960 -0.2385 -0.1227 -0.8035 0.4939
0.1898 -0.3528 0.9098 -0.1026 -0.0359
-0.5085 -0.6730 -0.0920 0.4778 0.2277
% Now V1 and V2 differ in sogn of 1st col (same w U1, U2), but still
>> norm(A-U2*S2*V2')
ans =
7.6825e-14
% and note that the diagonal matrices are the same.
>> S1
S1 =
127.0000 0 0 0 0
0 34.0000 0 0 0
0 0 3.5000 0 0
0 0 0 0.7200 0
0 0 0 0 0.0057
>> S2
S2 =
127.0000 0 0 0 0
0 34.0000 0 0 0
0 0 3.5000 0 0
0 0 0 0.7200 0
0 0 0 0 0.0057
% Ok, now let's really mess with the uniqueness of the SVD
% by setting two singular values to the same value.
>> S1 = diag([127, 34, 34, 0.72, 0.00573])
S1 =
127.0000 0 0 0 0
0 34.0000 0 0 0
0 0 34.0000 0 0
0 0 0 0.7200 0
0 0 0 0 0.0057
% Reform A from the new S1 and recompute the calculated SVD
>> A = U1*S1*V1';
>> [U2,S2,V2] = svd(A);
% By the way, what's the 2-norm of A?
>> norm(A)
ans =
127.0000
% The 2-norm of a matrix is its largest singular value!
% Compare U1 and U2
>> U1
U1 =
-0.1735 -0.2978 0.4339 0.8304 0.0575
-0.5919 -0.1608 -0.7489 0.2185 -0.1230
0.7291 0.1574 -0.4449 0.4549 -0.1970
-0.2783 0.7273 0.1992 0.1386 -0.5785
-0.1029 0.5760 -0.1147 0.1910 0.7798
>> U2
U2 =
-0.1735 0.2294 -0.4737 -0.8304 0.0575
-0.5919 -0.7302 0.2314 -0.2185 -0.1230
0.7291 -0.3084 0.3572 -0.4549 -0.1970
-0.2783 0.5334 0.5330 -0.1386 -0.5785
-0.1029 0.1858 0.5571 -0.1910 0.7798
diag(S1)'
ans =
127.0000 34.0000 34.0000 0.7200 0.0057
% Note that since sigma_2 = sigma_3, the 2nd and 3rd cols
% of U (and V) are indeterminate! The prescribed SVD and
% calculated SVD now differ in 2nd and 3rd cols of U and V
% Let's make three equal singular values
>> S1 = diag([127, 34, 34, 34, 0.00573])
S1 =
127.0000 0 0 0 0
0 34.0000 0 0 0
0 0 34.0000 0 0
0 0 0 34.0000 0
0 0 0 0 0.0057
>> A = U1*S1*V1';
>> [U2,S2,V2] = svd(A)
>> V1
V1 =
-0.6683 -0.0590 -0.1033 0.2860 0.6763
-0.4695 -0.6019 -0.3716 -0.1836 -0.4955
0.1960 -0.2385 -0.1227 -0.8035 0.4939
-0.1898 -0.3528 0.9098 -0.1026 -0.0359
0.5085 -0.6730 -0.0920 0.4778 0.2277
>> V2
V2 =
-0.6683 0 -0.3098 -0.0000 -0.6763
-0.4695 -0.7264 -0.0690 0.0405 0.4955
0.1960 -0.5303 0.6556 -0.0814 -0.4939
-0.1898 0.1663 0.3311 0.9085 0.0359
0.5085 -0.4042 -0.5999 0.4079 -0.2277
>> diag(S1)'
ans =
127.0000 34.0000 34.0000 34.0000 0.0057
% Notice now that the 2nd, 3rd, and 4th cols of V differ.
% But the 2nd, 3rd, and 4th cols of V1 span the same 3d subspace
% of R^5 and the same cols of V2. Can show this by expanding
% the 2nd (and 3rd and 4th) col of V1 in the 2,3,4 cols of V1
% Let x be the 2nd col of V1
>> x = V1(:,2);
% and v2,v3,v4 by the 2,3,4 cols of V2
>> v2 = V2(:,2);
>> v3 = V2(:,3);
>> v4 = V2(:,4);
% Show that x is in span(v2,v3,v4)
>> x
x =
-0.0590
-0.6019
-0.2385
-0.3528
-0.6730
% by expanding x in the 3d orthonormal set (v2,v3,v4)
>> (v2'*x)*v2 + (v3'*x)*v3 + (v4'*x)*v4
ans =
-0.0590
-0.6019
-0.2385
-0.3528
-0.6730
>> norm(x - ((v2'*x)*v2 + (v3'*x)*v3 + (v4'*x)*v4))
ans =
2.7336e-16
% Hooray! x (2nd col of V1) lies in the subspace spanned by the 2,3,4 cols of V2.
gibson/teaching/fall-2014/iam961/svddemo.txt · Last modified: 2014/09/25 12:30 by gibson
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