# Stuck at showing $U = V$ in SVD of Hermitian, positive definite matrices

Show $$U = V$$ in SVD of Hermitian, positive definite matrice $$A_{n \times n}$$

## Try

Since eigenvalues and singular values coincide in Hermitian matrices,

$$AV = U \Sigma = U \Lambda$$

where $$\Sigma$$ is the singular value diagonal matrix with decreasing diagonal, and $$\Lambda$$ is the eigenvalue diagonal matrix, i.e. $$\Sigma = \Lambda$$.

This implies that

$$Av_i = \lambda_i u_i \ \ (\ast)$$

where $$i=1,\cdots, n$$, which does not necessarily imply that $$v_i$$ are eigenvectors, and $$v_i = u_i$$, $$\forall i$$.

Let

$$v_i = c_{i1}\xi_1 + \cdots + c_{in}\xi_n \\ u_i = d_{i1}\xi_1 + \cdots + d_{in}\xi_n \\$$

where $$\xi_i$$ is the eigenvectors corresponding to $$\lambda_i$$.

The $$(\ast)$$ implies that, $$\forall i$$,

$$c_{i1}\lambda_1\xi_1 + \cdots + c_{in}\lambda_n\xi_n = d_{i1}\lambda_i\xi_1 + \cdots + d_{in}\lambda_i\xi_n$$

thus, since $$[\xi_1 | \cdots | \xi_n]$$ : invertible by spectral theorem,

$$\begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \cdots & c_{nn} \\ \end{bmatrix} \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{bmatrix} = \begin{bmatrix} d_{11} & d_{12} & \cdots & d_{1n} \\ d_{21} & d_{22} & \cdots & d_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ d_{n1} & d_{n2} & \cdots & d_{nn} \\ \end{bmatrix}^T \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n \end{bmatrix}$$

I notice that the matrices of $$c_{ij}$$ and $$d_{ij}$$ are invertible by the invertibility of $$U$$ and $$V$$, but I'm stuck at showing the identity.

Any help will be appreciated.

• $U$ contains the eigenvectors of $AA^H$ and $V$ has those of $A^HA$. Since $A$ is hermitian, $A^HA = AA^H$ implying that the matrices $U$ and $V$ are identical. – sudeep5221 May 29 at 1:29
• @sudeep5221 But when there exists $\sigma_i = \sigma_j$, $i \neq j$? – Moreblue May 29 at 3:57
• @sudeep5221 I don't see how your argument works. The equality $A^HA=AA^H$ is also satisfied by every normal matrix $A$, but unless $A$ is positive semidefinite, we cannot possibly have $U=V$ in the SVD of a normal matrix $A$. – user1551 May 29 at 5:02
• @user1551 Oh I see what you are trying to say. I had in mind what you wrote in your answer and took it for granted forgetting to point that out clearly. Sorry about that and thanks for pointing that out. :) – sudeep5221 May 29 at 16:56

Here is one way to prove it. Note that every positive semidefinite matrix $$P$$ has a unique positive semidefinite square root $$P^{1/2}$$. In fact, $$P^{1/2}=f(P)$$ where $$f$$ is the Lagrange interpolation polynomial that maps each eigenvalue of $$P$$ to its square root.
Now, in your case, since $$A^2=AA^\ast=US^2U^\ast$$, by the uniqueness of positive definite square root, we have $$A=USU^\ast$$. Hence $$U=V$$.