1

Question: Suppose $A$ is a complex $n\times n$ matrix and let $T:\mathbb{C}^{n\times n}\rightarrow\mathbb{C}^{n\times n}$ be the linear transofrmation given by $T(B)=AB+BA$ for $B\in\mathbb{C}^{n\times n}$. Prove that if $A$ is a nilpotent matrix, then $T$ is a nilpotent operator.

This problem is proved here (well, having $AB-BA$ instead of $AB+BA$, but who cares): $T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator., but I have a (elementary) question about the proof: Since \begin{equation}T(B)=AB+BA \implies T^2(B)=T(AB+BA), \text{by applying $T$ to both sides.} \\ = A(AB+BA)+(AB+BA)A \\ =A^2B+ABA+ABA+BA^2 \\ A^2B+2ABA+BA^2 \\\implies T^3(B)=T(A^2B+2ABA+BA^2), \text{by applying $T$ to both sides again} \\ =A(A^2B+2ABA+B^2A)+(A^2B+2ABA+B^2A)A \\ =A^3B+2A^2BA+B^2A^2+A^3B+2A^2BA+B^2A^2 \\ = 6A^3B+2A^2B^2 \end{equation}

So, if we proceed in this way, then since $A$ is nilpotent, we will reach a number (after applying $T$ so many times) where $A^m=0$, thus the terms in the expansion will always be zero after hitting that $m$, and so $T$ is a nilpotent linear operator.... is this the idea? Am I missing something? If that's true, then shouldn't the argument be done more formally using induction rather than just the first few cases?

Ben Grossmann
  • 225,327
User7238
  • 2,474
  • Eigenvalues of $T$ are the sum two eigenvalues of $A$. Since $A$ is nilpotent, the only eigenvalue of $A$ is zero. Thus, the only eigenvalue of $T$ is zero. – Sungjin Kim Sep 12 '21 at 21:04
  • @SungjinKim So what you mean is, if $\lambda$ is an eigenvalue of $A$, then as $A$ is nilpotent, say $A^m=0$, we have that $\lambda^m$ is an eigenvalue of $A$ and so $\lambda^m=0$. Thus $\lambda=0$... how does this imply that $T$ is a nilpotent operator? – User7238 Sep 12 '21 at 21:41
  • If $\mu$ is an eigenvalue of $T$, then an eigenvector of $T$ is a $\mathbb{C}[x]$-module homomorphism between $M_{A-\mu I}$ and $M_{-A}$ where $M_A$ is the $\mathbb{C}[x]$-module with $x v = Av$ for all $v\in M_A$. The only way that we can have a nonzero eigenvector of $T$ is to have $\lambda_1-\mu=-\lambda_2$ where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$. – Sungjin Kim Sep 12 '21 at 21:48
  • If $A$ is nilpotent, then $\lambda_1=\lambda_2=0$ is the only possibility. Thus, $\mu = \lambda_1+\lambda_2 = 0$. – Sungjin Kim Sep 12 '21 at 21:50

2 Answers2

3

There is a slightly nicer way to think about the calculations involved in this problem (which will lead to the formula from Ben Grossmann).Let $L_A:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ be left multiplication by $A$, and similarly let $R_A:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ be right multiplication by $A$. The clever observation is that $L_A$ and $R_A$ commute with each other, and that $T=L_A+R_A$. Since they commute, we can use the binomial theorem: $$T^k(B)=(L_A+R_A)^k(B)=\sum_{i+j=k}\binom{k}{i} L_A^iR_A^j(B)=\sum_{i+j=k}\binom{k}{i}A^i B A^j.$$

Then, if $k\geq 2m-1$ and $i+j=k$, then either $i\geq m$ or $j\geq m$, and since $A^m=0$ by assumption, each term in the sum vanishes.

Aaron
  • 24,207
  • How do we know that $L_A$ and $R_A$ commute with each other? Is that just the assumption in the problem? – User7238 Sep 12 '21 at 21:43
  • If you multiply on the left and then the right, you get ABA, but if you multiply on the right and then the left, you get ABA. – Aaron Sep 12 '21 at 21:44
  • Ahhh, I see. Thank you! – User7238 Sep 12 '21 at 21:50
  • 1
    It is worth noting that ANY left multiplication operator commutes with ANY right multiplation operator. $L_A R_C (B)=R_C L_A (B)=ABC$. – Aaron Sep 12 '21 at 22:12
1

Yes, that is the idea. And yes, a the presentation of this proof in the question was not rigorous.

One way to formalize the proof is to prove by induction that we have $$ T^k(B) = \sum_{j=0}^k \binom kj A^j BA^{k-j}. $$ From there, it suffices to note that if $A^m = 0$ and $k \geq 2m-1$, then for all $0 \leq j \leq k$ it holds that either $j \geq m$ or $k-j \geq m$, which means that each term in the sum is zero, which means that $T^k$ is indeed the zero transformation.

Ben Grossmann
  • 225,327