$AB-BA=A$, Then A is singular?

Question

Title is the question, I tried taking trace both side and got trace of $A$ is zero, now to conclude $A$ is singular, suppose $A$ is non singular, then multiplying both side by Inverse of $A$ we get $B$ is similar to $B+I$, which is impossible as similar matrices have same eigen values where as eigen values of $B+I$ is one unit more than the eigen values of $B$.

Is my solution make sense? Thanks for helping, I am using Android phone so not able to write mathematical symbols as it takes too much time to write.

Answer 1

Sure, your solution makes sense to me. In fact you can argue the same thing just from the trace: if $A$ is nonsingular then $B+I = ABA^{-1}$, and taking the trace of both sides gives $\mathrm{tr}(B)+n = \mathrm{tr}(B)$, which is impossible.

Answer 2

If $AB-BA = I$ then $trace(AB-BA) = trace(AB)-trace(BA) = trace(AB)-trace(AB)=0$ yet, $trace(I) = n$ for an $n \times n$ matrix. In short, it is not possible to have square matrices $A,B$ for which the commutator $[A,B] = AB-BA = I$. So, I'm not sure what the question means...

Answer 3

For two $n\times n$ matrices $A,B$ we cannot have $$ AB-BA = I $$ Since $$ \text{Tr}(AB) = \text{Tr}(BA) $$ and hence $ \text{Tr}(AB-BA)=0 $, while $\text{Tr}(I)=n$ (the size of the matrix)

Answer 4

In finite dimension the trace of a commutator is always zero.

$$ Tr(AB -BA) = Tr (AB) -Tr (BA) = \sum_{i,j=1}^{n} A_{ij}B_{ji} - \sum_{j,i=1}^n B_{ji}A_{ij} $$ In an infinite dimensional space this may not be the case, for example consider the space of smooth square integrable funcions $f(x)$ on $\mathbb{R}$, $L_{2}(\mathbb{R})$, and the linear operators (multiplication by) $x$ and (differentiation) $\partial =\frac{\partial}{\partial x}$, then

$$ [x, \partial] = id $$ which clearly does not have nil trace.

Answer 5

I see that the question was further modified. All this is unbearable.

About the equality $[A,B]=A$.

Case 1. The dimension is finite; $A,B\in M_n(K)$. According to the Jacobson's lemma, $[A,B]$ (and then $A$) is nilpotent. Moreover the previous result remains true when $K$ is a commutative ring with unity, under the condition that $n!$ is not a zero-divisor.

Case 2. The dimension of $E$ is infinite and $f,g\in L(E)$ are continuous. $fg-gf=f$ implies that, for every integer $k$, $f^kg-gf^k=kf^k$. Then $k||f^k||\leq 2||g||||f^k||$. Assume that, for every $k$, $f^k\not=0$; then, for every $k$, $k\leq 2||g||$, that is contradictory. Finally $f$ is nilpotent.