Is the Matrix Nilpotent?

Question

Let $A, B$ be nilpotent matrices over $\Bbb C$ where $AB=BA$. Are the following matrices nilpotent $$A-2B $$

$$A^3B$$

I don't really know where to start here. I saw a solution with the binomial expansion, however it seemed hard to understand.

Is there a different way?

Thanks,

Alan

Answer 1

Suppose that $A^a = 0$ and that $B^b=0$ for some integers $a,b\geq 1$ (i.e. that $A$ and $B$ are nilpotent), and furthermore that $AB=BA$ (i.e. that they commute).

You have then that $(A^3B)^{ab} = A^{3ab}B^{ab} = (A^a)^{3b}(B^b)^a = 0^{3b}0^a=0$

(Note: we could have used a smaller exponent than $ab$ and gotten a similar result, but it is unimportant. All we needed was to show that $A^3B$ is nilpotent, but we didn't care what the smallest such power is that works. An exponent as small as $\min\{a,b\}$ would have sufficed.)

(Also note: the step $(A^3B)^k = A^{3k}B^k$ is only valid if $A$ and $B$ commute. This is not true in general.)

As for $(A-2B)$, we approach similarly and use a weaker form of the binomial expansion argument that hopefully you will understand better than the solution you saw earlier.

$(A-2B)^{a+b} = \underbrace{(A-2B)(A-2B)(A-2B)\cdots (A-2B)}_{a+b~\text{number of times}}$

If you expand the right-hand side by "foiling", you will see that every term will be of the form $A^i(-2B)^j$ where $i+j=a+b$. You will have $i\geq a$ or $j\geq b$.

In the first case, $A^i$ will be zero then, and any matrix times a zero matrix will be zero. In the second case, $(-2B)^j$ will be zero, and the product will again be zero.

As a result, all terms in the expansion of $(A-2B)^{a+b}$ will be zero, and so $(A-2B)^{a+b}=0$, showing its nilpotence.

Answer 2

I try a solution without the binomial theorem. I suppose that we are working in $\mathbb{C}$.

Let $p\geq 1$ such that $A^p=0$. Let $\lambda $ an eigenvalue of $A-2B$, and $x\not = 0$ an eigenvector for the eigenvalue $\lambda$, ie $Ax-2Bx=\lambda x$. We have $A^p x=0$; hence there exists $h\geq 1$ such that $A^hx=0$ and $A^{h-1}x\not =0$. Multiplying by $A^{h-1}$, (and using that $AB=BA$) we get $-2BA^{h-1}x=\lambda A^{h-1}x$. This show that $y=A^{h-1}x$ is an eigenvector for $B$, with eignevalue $-\lambda/2$. As $B$ is nilpotent, we get $\lambda=0$. We have show that all eigenvalues of $A-2B$ are $0$, hence $A-2B$ is nilpotent.