If $A$ is positive semidefinite, then $A=USU^T$?

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?

• 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. – user1551 Oct 17 '18 at 4:33

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.

• Are you saying that it is not true that if A is positive semidefinite, A is always symmetric? – John Doe Oct 17 '18 at 3:54
• 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. – EdG Oct 17 '18 at 4:11