# Every finite rank operator on a Hilbert space can be written $\sum_{j=1}^n s_j|u_j\rangle\langle v_j|$ with $\{u_j\},\{v_j\}$ orthonormal

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.

• 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? Commented May 19, 2023 at 7:59
• 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. Commented May 19, 2023 at 8:01
• 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. Commented May 19, 2023 at 8:08
• @FrederikvomEnde Yes by Singular Value decomposition Commented May 19, 2023 at 10:11

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}

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

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.