Special basis for an automorphism of matrix group $M(n,\mathbb C)$

Question

Let $T$ be an automorphism of matrix group $M(n,\mathbb C)$ and let $e_1$, $\ldots$, $e_n$ be the standard basis of $\mathbb C^n$. Denote by $E_{ij}$ the matrix such that $(E_{ij})^i_j = 1$ and $(E_{ij})^k_l = 0$ if $(k,l) \neq (i,j)$. Then $E_{ij}$ realises isomorphism of $\langle e_j \rangle$ with $\langle e_i \rangle$, where $\langle e \rangle = \{ \lambda e \mid \lambda \in \mathbb C \}$. Now define $E_{ij}' = T(E_{ij})$. How to show that there exist basis $e_1'$, $\ldots$, $e_n'$ of $\mathbb C^n$ such that $E_{ij}'$ realises isomorphism of $\langle e_j ' \rangle$ with $\langle e_i ' \rangle$? (I can't use the fact that $T$ is in fact an inner automorphism).

Answer 1

I'm assuming that $T$ is an isomorphism of $M(n,\mathbb C)$ as an algebra (I don't think that what you look for is true if you only require $T$ to be a linear isomorphism).

The matrices $E'_{ij}$ also satisfy the "matrix-unit relations": $$ E'_{ij}E'_{kl}=T(E_{ij})T(E_{kl})=T(E_{ij}E_{kl})=\delta_{j,k}\,T(E_{il})=\delta_{j,k}\,E'_{il}. $$

As $T(E_{jj})^2=T(E_{jj}^2)=T(E_{jj})$, each $E'_{jj}$ is a non-zero (by the injectivity of $T$) projection. Let $e'_1$ be a unit vector in the rank of $E'_{11}$ (i.e. $E'_{11}=e'_1(e'_1)^*$). Now define $$ e'_j=E'_{j1}e'_1. $$ Then $$ E'_{ij}e'_j=E'_{ij}E'_{j1}e'_1=E'_{i1}e'_1=e'_i. $$ It remains to show that $e'_1,\ldots,e'_n$ is a basis. Suppose that $\sum_jc_je'_j=0$. Then, for each $k$, $0=T(E_{1k})\sum_jc_jT(E_{j1})e'_1=c_kT(E_{11})e'_1=c_ke'_1$; so $c_k=0$. This shows that $e'_1,\ldots,e'_n$ are linearly independent.