Let $A,B$ be a complex $n\times n$ matrices.Put $X=AB-BA$ .Assume that $A$ is diagonalizable and $AX=XA$.Show that $X$ equals the zero matrix.
It is a classic problem when we don't konw $A$ is diagonalizable and we can prove that $X$ is nilpotent.There are also many answers to it on this website.But I can not show that $X$ is the zero matrix when $A$ is diagonalizable. How should I use the additional condition?
Let $(\lambda,v)$ be any eigenpair of $A$. Then $$ Xv=(AB-BA)v=ABv-B(\lambda v)=(A-\lambda I)Bv. $$ Since $A$ is diagonalisable, its minimal polynomial $m(x)$ is a product of distinct linear factors. Thus we may write $m(x)=q(x)(x-\lambda)$, where $q$ is a monic polynomial such that $q(\lambda)\ne0$. Now, on one hand, we have $$ q(A)Xv=q(A)(A-\lambda I)Bv=m(A)Bv=0. $$ On the other hand, as $A$ commutes with $X$, $$ q(A)Xv=Xq(A)v=X(q(\lambda)v)=q(\lambda)Xv. $$ Therefore $q(\lambda)Xv=0$. As $q(\lambda)$ is nonzero, consequently we have $Xv=0$. Since this is true for any eigenvector of $A$ and $A$ has an eigenbasis, we conclude that $X=0$.
Assuming WLOG that $A$ is diagonal we have
$\Big\Vert X\Big\Vert_F^2 = \text{trace}\Big(X^*AB\Big)-\text{trace}\Big(X^*BA\Big)=\text{trace}\Big(AX^*B\Big)-\text{trace}\Big(X^*BA\Big) =0$
where $A$ being diagonal means it is normal so it commutes with $X^*$ per Fuglede and $\text{trace}\Big(AX^*B\Big)=\text{trace}\Big(X^*BA\Big)$ per cyclic property of trace.
After diagonalizing, we may assume $A$ is a diagonal matrix. Thus, we may write it in the block diagonal form as $A = \bigoplus_{i=1}^m c_iI_{k_i}$ where $c_i \in \mathbb{C}$, $c_i \neq c_j$ whenever $i \neq j$, and $k_i$ are positive integers s.t. $\sum_{i=1}^m k_i = n$. Then the commutant of $A$ is $\{A\}’ = \bigoplus_{i=1}^m M_{k_i}(\mathbb{C})$, which can be verified by a direct calculation. Thus, as $X \in \{A\}’$ we must have $X_{kl} = 0$ whenever $k, l \in \{1, \cdots, n\}$ are in different blocks. As $X = AB - BA$ and $A$ is diagonal, we have, when $k, l \in \{1, \cdots, n\}$ are in different blocks,
$$0 = X_{kl} = (A_{kk} - A_{ll})B_{kl}$$
Since $k, l$ are in different blocks, $A_{kk} \neq A_{ll}$, whence $B_{kl} = 0$ whenever $k, l$ are in different blocks. But this means $B \in \bigoplus_{i=1}^m M_{k_i}(\mathbb{C}) = \{A\}’$, i.e., $A$ and $B$ commute, so $X = AB - BA = 0$.
The map $T: M_n\big(\mathbb F\big)\longrightarrow M_n\big(\mathbb F\big)$ given by $T\big(C\big) = AC-CA$ is diagonalizable since $A$ is.
$T^2\big(B\big)=T\big(AB-BA\big)=T\big(X\big)=AX-XA=\mathbf 0$
$\implies B \in \ker T^2=\ker T$ since $T$ is diagonalizable, i.e. $\mathbf 0=T\big(B\big)=X$
Idea : by similarity you may assume that $A$ is a diagonal matrix with blocks $\lambda_iI_i$ where $\lambda_i$ are the eigenvalues and $I_i$ the $n_i\times n_i$ identity matrix. Then as $XA=AX$ the matrix $X$ has to be likewise block-diagonal with $n_i\times n_i$ blocks $X_i$ corresponding (in the same position) to the $\lambda_iI_i$ blocks in $A$. If you want $AB-BA$ to be not null you have to choose some non zero entries outside of the $n_i\times n_i$ blocks $B_i$ on the diagonal. But then $AB-BA$ has non-zero entries outside the $n_i\times n_i$ diagonal blocks hence contradiction with $XA=AX$. The easy case is when $A$ is $n\times n$ and has $n$ distinct eigenvalues, because then $AX=XA$ entails that $X$ is a diagonal matrix, and nilpotent hence null.
