I have a product of matrices that have the following form

$$ {\bf A} ^H {\bf A}$$

where subscript $H$ means hermitian transpose.

I am trying to find the eigen value decomposition (EVD) of ${\bf A^HA}= U D U ^H$ therefore I used MATLAB and function eig

$$[{\bf U, D}] = eig ({\bf A^HA})\tag 1$$

As a sanity check and in my opinion the product can be written as

$${\bf A^HA = UD^{1/2}V^{H} V D^{1/2} U}$$

where ${\bf UD^{1/2}V^{H}}$ is the SVD decomposition of A, so I used svd in MATLAB to obtain SVD of matrix ${\bf A}$

$${\bf [ U1,D,V] = svd(A)}\tag2$$

In my opinion ${\bf U}$ in (1) and ${\bf U1}$ in (2) should be equal but in MATLAB are not equal. Should they be equal or not?

  • 1
    $\begingroup$ The above suggests that $A=U_1 D V^*$, so $A^*A = V D^2 V^*$ (that is, look at $V$ not $U$, $U_1$). Also, the eigenvalues of $A^*A$ may not be ordered in the same way as the singular values. $\endgroup$
    – copper.hat
    Sep 1, 2015 at 4:52
  • $\begingroup$ thanks but do you agree that V and U should be matrix of same vector? $\endgroup$
    – Henry
    Sep 1, 2015 at 10:24

1 Answer 1


If $A = U D^{\frac{1}{2}}V^H$ is the SVD of $A$ then $$ A^HA = V D^{{1}{2}} U^H U D ^{\frac{1}{2}} V^H = V DV^H . $$ However, according to your definition, the spectral decomposition of $A^HA$ is given by $$ A^H A = U DU^H . $$ These are not equal. So you have a problem with your notation.


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