# eigenvalue of orthogonal matrix

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
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

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