这是一份liverpool利物浦大学PHYS202的成功案例

Therefore we have demonstrated how we can always construct a unitary representation by the transformation:
$$
\hat{A}{x}=d^{-1 / 2} U^{-1} A{x} U d^{1 / 2}
$$
where
$$
\begin{aligned}
&H=\sum_{x=1}^{h} A_{x} A_{x}^{\dagger} \
&d=\sum_{x=1}^{h} \hat{A}{x} \hat{A}{x}^{\dagger}
\end{aligned}
$$
and where $U$ is the unitary matrix that diagonalizes the Hermitian matrix $H$ and $\hat{A}{x}=U^{-1} A{x} U$.

PHYS202 COURSE NOTES ：

However, if $c=0$ then we cannot write but instead we have to consider $M M^{\dagger}=0$
$$
\sum_{k} M_{i k} M_{k j}^{\dagger}=0=\sum_{k} M_{i k} M_{j k}^{} $$ for all $i j$ elements. In particular, for $i=j$ we can write $$ \sum_{k} M_{i k} M_{i k}^{}=\sum_{k}\left|M_{i k}\right|^{2}=0
$$
Therefore each element $M_{i k}=0$ so that $M$ is a null matrix. This completes proof of the case $\ell_{1}=\ell_{2}$ and $M=\mathcal{O}$.

Finally we prove that for $\ell_{1} \neq \ell_{2}$, then $M=\mathcal{O}$. Suppose that $\ell_{1} \neq \ell_{2}$, then we can arbitrarily take $\ell_{1}<\ell_{2}$. Then $M$ has $\ell_{1}$ columns and $\ell_{2}$ rows. We can make a square $\left(\ell_{2} \times \ell_{2}\right)$ matrix out of $M$ by adding $\left(\ell_{2}-\ell_{1}\right)$ columns of zeros