If $\text{tr}(A) = \text{rank}(A) = 1$, find the Jordan Canonical Form of $A$.

Question

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.

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

Answer 1

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.

Answer 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 $1$.