# How does the matrix rank affects its Jordan Normal Form?

Let's say I have a matrix whose rank is $\rho(A) =1$.

How does it affect its Jordan Normal Form?

I know that a matrix A is similar to a Jordan Matrix and that means it has the same rank. So if a matrix has a rank of one, doesn't it mean that the Jordan Normal form is:

\begin{pmatrix} \lambda & 1 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}

A matrix whose rank is 1?

Alan

No, because the matrix you're showing is not a Jordan normal form at all (unless $\lambda=0$). Every off-diagonal $1$ need to appear inside a single Jordan block, that is, connecting two diagonal elements with the same value.
$$\left(\begin{array}{c|cc|c} \mathbf{0}_{a\times a} \\ \hline & 0 & 1 \\ & 0 & 0 \\ \hline &&& \mathbf{0}_{b\times b} \end{array}\right)$$ and $$\operatorname{Diag}(0,\ldots,0,\lambda,0,\ldots,0) = \left(\begin{array}{c|c|c} \mathbf{0}_{a\times a} \\ \hline & \lambda \\ \hline && \mathbf{0}_{b\times b} \end{array}\right) \text{ with }\lambda\ne 0$$