# Is $U=V$ in the SVD of a symmetric positive semidefinite matrix?

Consider the SVD of matrix $A$:

$$A = U \Sigma V^\top$$

If $A$ is a symmetric, positive semidefinite real matrix, is there a guarantee that $U = V$?

Second question (out of curiosity): what is the minimum necessary condition for $U = V$?

## 3 Answers

If the matrix is symmetric then $U=V$, as the by the spectral theorem we know that the eigenvalue decomposition and the singular value decomposition must be the same. From that we see that $U = U\Lambda U^{-1}=U\Lambda U^T=U\Sigma V^T$, and as by the theorem $\Sigma = \Lambda$ then $U=V$.

• your counter example is not symmetric – tibL Sep 15 '16 at 12:24
• That's why it does not hold that $U=V$ – Josu Etxezarreta Martinez Sep 15 '16 at 12:25
• in the question it says $A$ is symmetric... – tibL Sep 15 '16 at 12:26
• Ok, I didn't read that, I'll edit now – Josu Etxezarreta Martinez Sep 15 '16 at 12:27
• Why does $V \Sigma U^t = U \Sigma V^t$ tell you that $U = V$? – John Hughes Sep 15 '16 at 12:34

First of all see that $U$ and $V$ are not unique in general. However you might be able to find a relation between distinct SVD of a matrix $A$ and working with real matrix makes things easier.

For a general real $A$, let singular values of $A$ be non-zero. If $A=U_1\Sigma V_1^T$ and $A=U_2\Sigma V_2^T$ then from this link, there is a diagonal matrix $D=\mathrm{diag}(\pm 1,\dots,\pm 1)$ such that: $$U_1=U_2D, V_1=V_2D.$$ Now suppose that $A$ is a normal matrix with positive eigenvalues. It can be orthogonally diagonalized. Then we can see that: $$A=UDU^{T}$$ This is a SVD of $A$. So for $A=U_1\Sigma V_1^T$ then $U_1=UD$ and $V_1=UD$ which implies that $U_1=V_1$. In other words having normal matrix with positive eigenvalues is sufficient for having $U=V$. This class includes positive definite matrices. When zero singular values are permitted, then the situation is more tricky. Take zero matrix for instance.

Note the emphasis on being positive semi-definite. If $\mathbf A$ is singular, there is no such guarantee, and $\mathbf U$ and $\mathbf V$ can be different. As @Arash said, consider zero matrix, the SVD is not unique.

However, if we consider the column space or span of $\mathbf A$, and project $\mathbf U$ and $\mathbf V$ on this space, the projected U and V are equal.

It seems non-singularity also provides the necessary condition for $\mathbf U=\mathbf V$. But I need to double check this.