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


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 $v_i = u_i$, $\forall i$.


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

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.

  • 2
    $\begingroup$ $U$ 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. $\endgroup$
    – sudeep5221
    May 29 '19 at 1:29
  • $\begingroup$ @sudeep5221 But when there exists $\sigma_i = \sigma_j$, $i \neq j$? $\endgroup$
    – Moreblue
    May 29 '19 at 3:57
  • 1
    $\begingroup$ @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 $A$. $\endgroup$
    – user1551
    May 29 '19 at 5:02
  • $\begingroup$ @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. :) $\endgroup$
    – sudeep5221
    May 29 '19 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 $U=V$.


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .