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i have following question and my suppose about it and please tell me if i am wrong or not,if we take some random matrix

A=rand(3,3)

A =

    0.8147    0.9134    0.2785
    0.9058    0.6324    0.5469
    0.1270    0.0975    0.9575

and make it's SVD [U E V]=svd(A)

U =

   -0.6612   -0.4121   -0.6269
   -0.6742   -0.0400    0.7375
   -0.3290    0.9103   -0.2513


E =

    1.8168         0         0
         0    0.8389         0
         0         0    0.1815


V =

   -0.6557   -0.3056    0.6904
   -0.5848   -0.3730   -0.7204
   -0.4777    0.8761   -0.0658

i will get orthogonal matrices,now if i do SVD on each matrix,for example on U

[U1 E1 V1]=svd(U)

U1 =

   -0.0882   -0.6612    0.7450
   -0.3656   -0.6742   -0.6417
    0.9266   -0.3290   -0.1823


E1 =

    1.0000         0         0
         0    1.0000         0
         0         0    1.0000


V1 =

         0    1.0000         0
    0.8944         0   -0.4472
   -0.4472         0   -0.8944

i got diagonal matrix with entries $1$,i as thinking if why it is so?does it work only for orthogonal matrces got by SVD decomposition or in general every orthogonal matrix satisfy it?i remembered that basci idea of eigenvalue using geometrical interpretation is

to compress/stretch/change direction of vector

$A*x=\lambda *x$

because orthogonal matrix preserves length(it is isometric transformation)maybe that why every orthogonal matrix has eigenvalues as $1$,am i right?thanks in advance

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compare: [ wolframalpha.com/input/… ] –  janmarqz Jan 17 at 17:56
    
i am taking SVD not on original matrix,but on left side matrix,on U –  dato datuashvili Jan 17 at 18:00
1  
there in WA you can continue to experiment –  janmarqz Jan 17 at 18:03
    
of course we are studying from experiment :) –  dato datuashvili Jan 17 at 18:07

1 Answer 1

up vote 2 down vote accepted

The singular values of $A$ (i.e. the diagonal values of $E$ given in the singular value decomposition) are the square roots of the non-zero eigenvalues of $A\cdot A^*$. see: http://en.wikipedia.org/wiki/Singular_value_decomposition

The singular values of a unitary matrix are thus the square roots of its eigenvalues. But the eigenvalues of a unitary matrix are all 1.

In other words: you are correct that an orthogonal matrix is an isometry so that all of its eigenvalues are 1 and this is why its singular values are 1.

share|improve this answer
1  
thank you very much my friend –  dato datuashvili Jan 17 at 18:07
    
so it works on every orthogonal matrices without SVD right? –  dato datuashvili Jan 17 at 18:14
    
@datadatuashvili Yes. For the same reason. –  gabe Jan 17 at 20:03
1  
thanks and wish you everything best in your live –  dato datuashvili Jan 17 at 20:12

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