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

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$A$ 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.

## marked as duplicate by Martin R, Mostafa Ayaz, José Carlos Santos, Saad, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 5 '18 at 0:27

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

• Can't this be better as a comment? – Amin235 Mar 4 '18 at 20:34

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 $AB - BA \ne 0$.

• You are assuming the matrices are all $2 \times 2$. – angryavian Mar 4 '18 at 21:08
• @angryavian Thanks, that $2 \times 2$ in the question has fooled me. – mechanodroid Mar 4 '18 at 21:19