Let $T$ be a finite rank operator on a Hilbert space $H$. Then there exist orthonormal sets $\{u_j\},\{v_j\}$ and positive scalars $\{s_j\}$, with $j=1,2,\ldots,n$, such that $$T=\sum\limits_{j=1}^n s_j|u_j\rangle\langle v_j|.$$ Here $|u\rangle \langle v|(x)=\langle v,x\rangle u$

Given finite rank operator $T$, $T$ defines a linear correspondence between the finite dimensional spaces $\text{Ker}(T)^\perp$ and $\text{Ran}(T)$. Let $\{u_1,u_2,\ldots, u_n\}$ be an orthonormal basis for $\text{Ran}(T)$. Then there is $\{v_1,\ldots,v_j\}\in\text{Ker}(T)^\perp$ such that $Tv_j=u_j$ for $j=1,\ldots,n$. As $T(v_j/\lVert v_j\rVert)=(1/\lVert v_j\rVert) u_j$, we may assume without loss of generality that $Tv_j=s_ju_j$ and $\lVert v_j\rVert=1$.

But this $\{v_j\}$ may not be an orthonormal set. If it is, we are done. If we apply Gram Schmidt orthogonalization on $\{v_j\}$, the $\{u_j\}$ will be changed and may not be orthonormal set any more.

Can anyone suggest a way out? Thanks in advance for your help.

  • 1
    $\begingroup$ Can you prove this when adding the assumption that $T$ is self-adjoint (so $T=\sum_j s_j|u_j\rangle\langle u_j|$)? If so, can you connect this special case to your original question using a suitably chosen unitary? $\endgroup$ Commented May 19, 2023 at 7:59
  • $\begingroup$ Why does it matter if the $u_i$'s are orthonormal? I would have thought that we only need the $v_j$'s to be orthonormal for your construction to work. Whether the $u_i$'s are orthonormal is immaterial. $\endgroup$
    – Kenny Wong
    Commented May 19, 2023 at 8:01
  • $\begingroup$ Your problem is equivalent to writing $T=\sum\limits_{j=1}^n|x_j\rangle\langle y_j|$ with $\{x_j\},\{y_j\}$ only orthogonal. $\endgroup$ Commented May 19, 2023 at 8:08
  • $\begingroup$ @FrederikvomEnde Yes by Singular Value decomposition $\endgroup$ Commented May 19, 2023 at 10:11

2 Answers 2


Wlog, $H=\Bbb R^n$ and $T$ is bijective. Let $(v_j)$ be an orthonormal eigenbasis for $T^*T.$ Define $s_j>0$ by $T^*T(v_j)=s_j^2v_j,$ and let $u_j:=\frac1{s_j}T(v_j).$ Then $$\begin{align}\langle u_i,u_j\rangle&=\frac1{s_is_j}\langle Tv_i,Tv_j\rangle\\&= \frac1{s_is_j}\langle T^*Tv_i,v_j\rangle\\&=\frac1{s_is_j}\langle s_i^2v_i,v_j\rangle\\&=\frac{s_i}{s_j}\delta_{i,j}\\&=\delta_{i,j}. \end{align}$$


Apriori, your operator looks like

$$T = \sum_{kl} a_{kl} |e_k\rangle \langle f_l \mid $$

where $(\mid e_k\rangle)_{1}^n$, $(\mid f_f\rangle)_{1}^n$ are orthonormal system of finite size $n$. Now, to get our expression, we need to get the SVD of the matrix $(a_{kl})$. That is the idea.


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