0

Let $A \in M_{n \times n}(\mathbb{C})$ with $n > 1$. If $\text{tr}(A) = \text{rank}(A) = 1$, find the Jordan Canonical Form of $A$.

Since $A$ is a complex matrix, it must have a Jordan Form. Moreover, the number of Jordan block $J_*(0)$ with diagonal $0$ is $n - \text{rank}(A) = n - 1$. Hence, $A \sim \text{diag}\{J_{s_1}(\lambda_1), \lambda_{s_2}(0), \cdots, J_{s_{n}}(0) \}$, where $J_s(\lambda)$ denotes a Jordan Block with size $s$ and diagonal $\lambda$.

Note that $s_1 + \cdots + s_{n} = n$ and $s_j \geq 1$ for each $j$, we have that $s_j = 1$ for all $j$. Therefore, $$A \sim \text{diag}\{J_{1}(\lambda_1), J_1(0), \cdots, J_{1}(0) \}$$

Then $\lambda_1 = 1$ is given by the trace condition.

Thank @xbh for helpful comments and @peek-a-boo for his great answer. I made some edits to my original post above.

Rodrigo de Azevedo
  • 23,223
mathdoge
  • 986
  • 1
    What is the rank of each $J_{s_j} (\lambda_j)$? Then what is the minimal rank of $A$ according to these Jordan blocks? – xbh Jun 12 '19 at 02:34
  • @xbh If $\lambda_j \neq 0$, then the rank of $J_{s_j}(\lambda_j)$ should be $s_j$; if $\lambda_j = 0$, then the rank should be $s_j - 1$. So accordingly, $\lambda_1 = 0$, and only one $\lambda_j \neq 0$ for $2\leq j \leq n-1$, say $\lambda_2 \neq 0$ otherwise the rank condition would be violated. Using the trace condition, we know that $\lambda_2 = 1$. Is that correct? – mathdoge Jun 12 '19 at 02:55
  • Basically correct [if i read it correctly]. So what is the final form? – xbh Jun 12 '19 at 02:57
  • Well… then $\DeclareMathOperator \rank {rank} \DeclareMathOperator \diag {diag}\rank A = 2$, not $1$, since $\rank (J_2 (0)) = 1$… My answer is $\diag (1, \boldsymbol O_{n-1})$. Maybe the assertion "$# $ of Jordan blocks" is wrong… It seems to be "$#$ of Jordan blocks $J_*(0)$ is $n-1$" [if I remembered this correctly]. – xbh Jun 12 '19 at 03:08
  • @xbh Yes, you are definitely correct. I remember it wrong. – mathdoge Jun 12 '19 at 03:10

2 Answers2

7

I think you're overthinking it slightly with all the indices; sure the statements you made seem to be true, but you haven't fully exploited the rank condition imposed on $A$.

Since $\text{rank}(A) = 1$, we have that $\dim \ker(A)= n-1 > 0$. In words, this says $0$ is an eigenvalue of $A$, whose eigenspace is $n-1$ dimensional. Hence, in the JCF of $A$, there are guranteed to be $n-1$ zeros on the diagonal, which are $1 \times 1$ blocks. The last eigenvalue $\lambda$ has to be such that \begin{equation} \lambda + \underbrace{0 + \dots + 0}_{n-1 \text{ zeros}} = 1 = \text{tr}(A) \end{equation} Hence, the last eigenvalue has to be $1$ (and clearly it has to be a $1\times 1$ block). Thus, the JCF is a diagonal matrix with a single $1$, and $n-1$ zeros on the diagonal.

peek-a-boo
  • 65,833
2

Rank of $A$ is one implies $A:=uv^t$ for some $u,v \in \Bbb C^n$. So eigenvlaues of $A$ are $0$ and $1$(=trace) with multiplicity $n-1$ and $1$ respectively. So there are $n-1$ blocks corresponding to $0$ and one block corresponding to $1$.

Chinnapparaj R
  • 11,989