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Suppose we have an M$\times$N complex matrix $H$ and its singular value decomposition $H=U\Lambda V^*$ and an N$\times$N covariance matrix $R_s$ with its eigendecomposition $R_s = U_s\Lambda_sU_s^*$. We also have the eigendecomposition of $HR_sH^*$ as $U_A\Lambda_AU_A^*$. In my research problem setting, I know $U_A$, $H$ but not $R_s$ and $\Lambda_A$. I want to find $U_s$ using $H$ and $U_A$. I was wondering if there exists any connection between them.

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  • $\begingroup$ When you say that you don't know $R_d$, do you mean $R_s$? (Also, does "solve $U_s$" mean "find $U_s$"?) $\endgroup$
    – LSpice
    Jun 8, 2018 at 18:41
  • $\begingroup$ Yes, you are right and I have corrected my wording based on what you suggested. $\endgroup$
    – Jiawei Liu
    Jun 8, 2018 at 18:51
  • $\begingroup$ @Jiawei Liu Does my answer adequately answer your question? $\endgroup$ Jun 10, 2018 at 11:41
  • $\begingroup$ @Jiawei Liu Does my answer address your question? $\endgroup$ Jun 12, 2018 at 10:54
  • $\begingroup$ It does address my question. Appreciate it a lot @Mark L. Stone $\endgroup$
    – Jiawei Liu
    Jun 12, 2018 at 22:03

1 Answer 1

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$U_s$ is not recoverable from $H$ and $U_A$.

Consider the following examples, one each for $M < N, M > N, M = N$. In each example, there are 2 different $R$'s, $R1$ and $R2$, having different $U_s$'s while having the same $U_A$.

MATLAB output for $M=2, N=3$ example:

>> disp(H)
     1     0     0
     0     2     0
>> disp(R1)
     2     1     1
     1     2     1
     1     1     2
>> [U_s_R1,lambda_R1]=eig(R1)
U_s_R1 =
   0.408248290463863   0.707106781186547   0.577350269189626
   0.408248290463863  -0.707106781186547   0.577350269189625
  -0.816496580927726                   0   0.577350269189626
lambda_R1 =
   0.999999999999999                   0                   0
                   0   1.000000000000000                   0
                   0                   0   3.999999999999999
>> disp(R2)
   9.999999999999929   4.999999999999965   0.984522053823275
   4.999999999999965   9.999999999999929   0.984522053823275
   0.984522053823275   0.984522053823275   9.999999999999929
>> [U_s_R2,lambda_R2]=eig(R2)
U_s_R2 =
   0.707106781186548   0.177730756491407   0.684406150028616
  -0.707106781186547   0.177730756491407   0.684406150028616
                   0  -0.967896459542024   0.251349246280978
lambda_R2 =
   4.999999999999966                   0                   0
                   0   9.638432711095330                   0
                   0                   0  15.361567288904492
>> [U_A_R1,lambda_A_R1]=eig(H*R1*H')
U_A_R1 =
  -0.957092026489053   0.289784148688430
   0.289784148688430   0.957092026489053
lambda_A_R1 =
   1.394448724536011                   0
                   0   8.605551275463990
>> [U_A_R2,lambda_A_R2]=eig(H*R2*H')
U_A_R2 =
  -0.957092026489053   0.289784148688430
   0.289784148688430   0.957092026489053
lambda_A_R2 =
   6.972243622680004                   0
                   0  43.027756377319641

As can be seen, U_A_R1 = U_A_R2, but U_s_R1 shares only one column with U_s_R2, i.e., R1 has only one eigenvector in common with R2.


MATLAB output for $M=3, N=2$ example

>>  disp(H)
     1     0
     0     2
     1     1
>> disp(R1)
     1     1
     1     2
>>  [U_s_R1,lambda_R1]=eig(R1)
U_s_R1 =
  -0.850650808352040   0.525731112119133
   0.525731112119133   0.850650808352040
lambda_R1 =
   0.381966011250105                   0
                   0   2.618033988749895
>>  disp(R2)
  10.000000000000000   2.500000000000000
   2.500000000000000  10.000000000000000
>> [U_s_R2,lambda_R2]=eig(R2)
U_s_R2 =
  -0.707106781186547   0.707106781186547
   0.707106781186547   0.707106781186547
lambda_R2 =
   7.500000000000000                   0
                   0  12.500000000000000
>>  [U_A_R1,lambda_A_R1]=eig(H*R1*H')
U_A_R1 =
   0.666666666666668   0.711452386093986   0.222240990541188
   0.333333333333332  -0.551270143087020   0.764846467096310
  -0.666666666666666   0.435817314550478   0.604664224089342
lambda_A_R1 =
   0.000000000000002                   0                   0
                   0   0.675444679663242                   0
                   0                   0  13.324555320336762
>> [U_A_R2,lambda_A_R2]=eig(H*R2*H')
U_A_R2 =
   0.666666666666667   0.711452386093987   0.222240990541187
   0.333333333333333  -0.551270143087020   0.764846467096309
  -0.666666666666666   0.435817314550477   0.604664224089342
lambda_A_R2 =
   0.000000000000005                   0                   0
                   0  13.782917548737156                   0
                   0                   0  61.217082451262840

As can be seen, U_A_R1 = U_A_R2, but U_s_R1 doesn't share any columns with U_s_R2, i.e., R1 has no eigenvectors in common with R2.


MATLAB output for $M=2, N=2$ example

>> disp(H)
     1     0
     0     2
>> disp(R1)
     1     1
     1     2
>>  [U_s_R1,lambda_R1]=eig(R1)
U_s_R1 =
  -0.850650808352040   0.525731112119133
   0.525731112119133   0.850650808352040
lambda_R1 =
   0.381966011250105                   0
                   0   2.618033988749895
>>  disp(R2)
   9.999999999998895   4.285714285713818
   4.285714285713818   9.999999999998892
>> [U_s_R2,lambda_R2]=eig(R2)
U_s_R2 =
   0.707106781186547  -0.707106781186548
  -0.707106781186548  -0.707106781186547
lambda_R2 =
   5.714285714285076                   0
                   0  14.285714285712711
>>  [U_A_R1,lambda_A_R1]=eig(H*R1*H')
U_A_R1 =
  -0.966499648764670   0.256667935157024
   0.256667935157024   0.966499648764670
lambda_A_R1 =
   0.468871125850725                   0
                   0   8.531128874149275
>> [U_A_R2,lambda_A_R2]=eig(H*R2*H')
U_A_R2 =
  -0.966499648764669   0.256667935157025
   0.256667935157025   0.966499648764669
lambda_A_R2 =
   7.723733396502248                   0
                   0  42.276266603492211

As can be seen, U_A_R1 = U_A_R2, but U_s_R1 doesn't share any columns with U_s_R2, i.e., R1 has no eigenvectors in common with R2.

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  • $\begingroup$ I think your answer makes total sense to me, Mark! Thanks for your input! $\endgroup$
    – Jiawei Liu
    Jun 12, 2018 at 22:00

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