# $T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator.

Let $V$ be a vector space of $n\times n$ matrices over a field F, and let $A$ be a fixed $n\times n$ matrix. $T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator.

I have done in a way that $T^2(B)= T(AB-BA)= A^2B-2ABA+BA^2\Rightarrow T^3(B)=A^3B-3A^2BA+3ABA^2-BA^3$ Proceeding in this way since A is nilpotent $\exists m\in\Bbb N$ s.t $T^m(B)=0$.

Hence T is also a nilpotent operator. This a problem of Hoffman Kunze of Primary Decomposition Chapter. So can anyone give me any other solution because this solution depends on some sort of intution. Please don't use any Rational and Jordan Forms formula.

• Ok but how? Calculating the matrix is much laborious. I have done it for $2\times2$ case. If u have any idea u can write in the answer. – user152715 Sep 19 '14 at 17:18
• No, my question is different. – user152715 Sep 19 '14 at 17:28
• No, its not. There may be some steps in between this u can see but I can't. If u derive the matrix of T for $2\times2$ case. U will see the matrix of T does directly depend upon the elements of A but not directly as the structure of the matrix A. So the eigenvalue they have derived can't be derived in that way. Anyway if u still thinks that there is a relation then post 1 answer deriving the eigenvalue. – user152715 Sep 19 '14 at 17:41
• I'm getting confused between my As and Bs. – copper.hat Sep 19 '14 at 17:53
• Give the answer reading the question – user152715 Sep 19 '14 at 18:32

This is in the same spirit to your proof, but presented in a different way. If $\lambda B = AB-BA$ for some $B\ne0$ and some $\lambda$ in the algebraic closure of $F$, then $(A-\lambda I)B=BA$ and $(A-\lambda I)^k B=BA^k$ for any $k\ge1$. In particular, $(A-\lambda I)^nB=0$. However, if $\lambda$ is nonzero, $A-\lambda I$ would be invertible and hence $B=0$, which is a contradiction.

• Can U explain the line from $(A-\lambda I)B=BA$ to $(A-\lambda I)^kB=BA^k$. – user152715 Sep 20 '14 at 3:45
• @user152715 Try to prove by mathematical induction. E.g. $(A-\lambda I)^2B=(A-\lambda I)\left[(A-\lambda I)B\right]=(A-\lambda I)BA=\left[(A-\lambda I)B\right]A=BA^2$. – user1551 Sep 20 '14 at 8:38
• Why $A-\lambda I$ would be invertible for nonzero $\lambda$? – Majid May 20 '18 at 20:19
• @Majid As $A$ is nilpotent, all eigenvalues of $A-\lambda I$ are equal to $-\lambda\ne0$. Hence $A$ is invertible. Alternatively, as $A$ is nilpotent, $A^m=0$ for some $m\ge1$. So, if $(A-\lambda I)x=0$, then $0=A^mx=\lambda^m x$ and hence $x=0$. Therefore $\ker A=0$ and $A$ is invertible. – user1551 May 20 '18 at 20:27
• @user1551 Thanks! I got. – Majid May 20 '18 at 20:29

This is not an answer as it depends on certain characteristics of the underlying field (also, I haven't used the fact that $A$ is nilpotent).

Suppose $T(B) = \mu B$, that is $\mu B = AB-BA$. Then $\mu B^2 = BAB-B^2A= (AB-\mu B)B-B^2 A= AB^2 -B^2A - \mu B^2$, or $2 \mu B^2 = AB^2 -B^2 A = T(B^2)$.

So, if $\mu$ is an eigenvalue corresponding to an eigenvector $B$ of $T$, then $B^2$ is an eigenvector corresponding to the eigenvalue $2 \mu$. Hence $2^k \mu$ are eigenvalues.

Since there are only a finite number of eigenvalues (this is where I am making presumptions about the field), we have $\mu = 0$.

Your proof seems correct to me. This result is used for Engel's theorem in the theory of Lie algebras. If $x\in \mathbb{gl}(V)$ is nilpotent, then also $ad(x)$ is nilpotent, where $ad(x)(y)=[x,y]=xy-yx$ for $x,y\in \mathfrak{gl}(V)$. Indeed, $ad(x)^m$ is a linear combination of terms $x^iyx^{m-i}$.

I proved this the same way the OP proved this problem (coincidentally, I had the same problem as homework, proved it, and then went online to see if another proof was available, to verify my work -- as I worked all the way up to the expression [I'm guessing the OP did this as well]). $T^{m}(B)={\displaystyle{\sum\limits_{i=0}^{m}(-1)^{i}\binom{m}{i}A^{m-i}BA^{i}}}$, then factor $B$ out of the sum, which gives the result, for any $B\in\mathcal{M}_{n\times n}(F)$ -- etc. Further, I did happen to find another proof as well. I'll give a proof sketch, since, to me, this alternative proof is convoluted a little, or somewhat overkill; mine/OP's proof is shorter, less complicated, etc. Nevertheless, here is the proof-sketch.

Proof-Sketch: First, if $V$ is an $F$-vector space, we claim that if $T_{1},T_{2}\in\mathcal{L}(V,V)$ are nilpotent, commuting operators, then $T_{1}+T_{2}$ is nilpotent.

To prove the claim, we have $T_{1}^{k}=0_{\mathcal{L}(V,V)}=T_{2}^{m}$ for some $k,m\in\mathbb{N}$ since these operators are assumed nilpotent (suppose without loss of generality that $m\leq k$ and note that $0_{\mathcal{L}(V,V)}$ is the zero operator/element of $\mathcal{L}(V,V)$ which is my preference in notation). Use the Binomial Theorem to expand $(T_{1}+T_{2})^{k+m}$, and show that $T_{1}^{k+m-i}T_{2}^{i}=0_{\mathcal{L}(V,V)}$ for all $i=0,1,...,m,m+1,...,k+m$; note that the use of the Binomial Theorem in this case is possible since the operators commute.

Returning to the overall proof, we have $V=\mathcal{M}_{n\times n}(F)$ whenever $n\in\mathbb{N}$ is arbitrarily fixed, and we take $A\in\mathcal{M}_{n\times n}(F)$ arbitrarily also. Define the operators $T_{1},T_{2}\in\mathcal{L}(V,V)$ such that for all $B\in\mathcal{M}_{n\times n}(F)$ we have $T_{1}(B)=AB$ and $T_{2}(B)=-BA$. This all being said, we first need to show that on $\mathcal{M}_{n\times n}(F)$ we have:

(i) $T_{1}T_{2}=T_{2}T_{1}$ (i.e., $T_{1},T_{2}$ commute); and

(ii) $T_{1}+T_{2}=T$ (where $T$ is defined the OP's problem-statement - also, this follows trivially by the definition of a sum of operators).

Thus, to show $T$ is nilpotent, it is sufficient to show that $T_{1},T_{2}$ are both nilpotent operators on $\mathcal{M}_{n\times n}(F)$. This can be accomplished by claiming for any $r,s\in\mathbb{N}$, we have $T_{1}^{r}(B)=A^{r}B$ as well as $T_{2}^{s}(B)=(-1)^{s}BA^{s}$ and proceeding by induction on $r$ and $s$ in $\mathbb{N}$, respectively. Thus, since $T_{1},T_{2}$ are nilpotent, we apply the claim at the beginning of the proof to get the overall result, and we are done.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$

Like I said, to me this seems to be a bit much, since we have to show a bit more that any of the other proofs posted, but I decided to post it since the original claim at the beginning of the proof can be useful (note that claim is independent of the dimension of $V$). Hopefully there aren't any mistakes, as I tried to keep the above proof-sketch a little abstract so one who decides to use it can work out the details himself/herself (and I was a little quick about things too...sorry about that and that this is a bit lengthy still yet).

• To critique my own work, where I say $T^{m}(B)={\displaystyle{\sum\limits_{i=0}^{m}\binom{m}{i}A^{m-i}BA^{i}}}$, I think I need $T^{m}(B)={\displaystyle{\sum\limits_{i=0}^{m}(-1)^{i}\binom{m}{i}A^{m-i}BA^{i}}}$ instead (as it is mentioned above, $T$ is defined in the OP's problem-statement)? Going to edit this in now, awaiting any corrections, thoughts, etc. – Procore Jul 12 '18 at 23:55

Note that $$T=L-R \qquad L=\textrm{Multiply X at Left}\qquad R=\textrm{Multiply X at right}$$ Since $$L\circ R=R\circ L\qquad R^{n}=0\qquad L^{n}=0$$ if $A^n=0$, then $T^{2n}=0$ by binomial theorem.