$A$ and $B$ are square complex matrices of the same size and $\text{rank}(AB − BA) = 1.$ Show that $(AB − BA)^2= 0.$

Question

$A$ and $B$ are square complex matrices of the same size and $\text{rank}(AB − BA) = 1.$ Show that $(AB − BA)^2= 0.$

I am trying to understand the solution given here:

Let $C = AB − BA.$ Since $\text{rank} C = 1$, at most one eigenvalue of $C$ is different from $0.$ Also $\text{tr} C = 0$, so all the eigevalues are zero. In the Jordan canonical form there can only be one $2 × 2$ cage and thus $C^2 = 0.$

I do not understand the argument in the bold part. I know that every matrix with complex entries can be put into the Jordan Canonical form but why is there only one $2\times 2$ cage? Please explain. I am still learning this concept.

Answer 1

If there is a larger cage, then the rank of $C$ would be $> 1$.

Answer 2

Recall that the size of the largest Jordan block pertaining to the eigenvalue $\lambda$ is precisely the power of $(x - \lambda)$ in the minimal polynomial. Since $\sigma(AB - BA) = \{0\}$, the minimal polynomial of $AB - BA$ is of the form $x^k$, where $1 \le k \le n$.

Also, recall that the nullity ($\operatorname{null} T = \dim\ker T$) of a matrix $T$ is equal to the number of Jordan blocks in the Jordan form of $T$. To see this, notice that the nullity of every Jordan block is exactly one.

Now, using the Rank-Nullity theorem, we obtain:

$$n = \dim V = \operatorname{rank}(AB - BA) + \operatorname{null}(AB - BA) = 1 + \operatorname{null}(AB - BA)$$ implying $\operatorname{null}(AB - BA) = n - 1$. Therefore, the Jordan form has exactly $n-1$ blocks.

Since $n$ is the sum of sizes of all blocks, and blocks have size $\ge 1$, it must be that exactly $n-2$ blocks are of size $1 \times 1$, and one is of the size $2 \times 2$ .

Therefore, the size of the largest block is $2 \times 2$, so we conclude that the minimal polynomial is precisely $x^2$, implying $(AB - BA)^2 = 0$ and $AB - BA \ne 0$.