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If A is a positive semidefinite matrix, then it has a singular value decomposition $A=USV^T$ with $U=V$. My textbook states this as fact, but I cannot seem to prove it. Additionally, $S$ must have the square roots of the eigenvalues of $AA^T$ as diagonal elements, since $S$ is the matrix of singular values.

Edit: This would be trivial if all positive semidefinite $A$ are symmetric. In this case, $AA^T=A^2$ would have eigenvalues $\lambda^2$, where $\lambda$ are the eigenvalues of $A$, which are nonnegative by definition, so then $S=D$, the diagonal matrix of eigenvalues. Additionally, symmetric matrices are orthogonally diagonalizable, so we could write them as $A=UDU^T$, where $U$ have columns as the orthonormal eigenvectors. However, I do not believe this is true however, are positive semidefinite $A$ always symmetric?

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    $\begingroup$ This is a matter of definition. In mathematics, when $A$ is real, apart from the requirement that $x^TAx\ge0$ for all real vectors $x$, almost all textbooks also require that $A$ is symmetric in the definition of positive semidefiniteness. In other scientific disciplines, symmetry is sometimes not required. Your textbook most likely employs the mainstream definition and requires that $A$ is symmetric. $\endgroup$ – user1551 Oct 17 '18 at 4:33
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You are right and the textbook is wrong. If $A$ is symmetric, then $A$ has a singular value decomposition $A=USV^T$ with $V=U$. However, when $A$ is not symmetric, this is not the case.

A very simple example to illustrate that the textbook is wrong: \begin{align} A=USU^T= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} s_1 & 0 \\ 0 & s_2 \end{bmatrix} \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} a^2s_1+b^2s_2 & acs_1+bds_2 \\ acs_1+bds_2 & c^2s_1+d^2s_2 \end{bmatrix}. \end{align} As you can see, the matrix $A$ has to be symmetric.

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  • $\begingroup$ Are you saying that it is not true that if A is positive semidefinite, A is always symmetric? $\endgroup$ – John Doe Oct 17 '18 at 3:54
  • $\begingroup$ Yes, that's what I said (implicitly). Positive semidefinite means that for any $x$, we have $x^TAx \geq 0$. As an example, consider a skew matrix $A$, such that $A^T=-A$. It is easy to see that $x^TAx=0$ for all $x$, hence $A$ is a (somewhat special) positive semidefinite matrix. $\endgroup$ – EdG Oct 17 '18 at 4:11

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