Does this property hold for the Jordan form of a Matrix?

Question

Suppose $\lambda$ is an eigenvalue of the matrix $M$ with geometric multiplicity $r$.

I have the feeling that in the Jordan form of $M$, there are $r$ columns that contains only $\lambda$ (and no $1$) (there may be other columns that contain $\lambda$ but will also have $1$). Is this true?

Answer 1

Let $\pmb{M} \in \mathbb{C}^{N \times N}$ be a matrix. There exists an invertible matrix $\pmb{P}$ such that $\pmb{J }= \pmb{P}^{-1}\pmb{M}\pmb{P}$, under the following block diagonal form \begin{equation} \pmb{J} = \begin{bmatrix} \pmb{J}_1 & & & \\ & \pmb{J}_2 & & \\ & & \ddots & \\ & & & \pmb{J}_p\\ \end{bmatrix} \in \mathbb{C}^{N\times N} \end{equation} where \begin{equation} \pmb{J}_k = \begin{bmatrix} \lambda_k & 1& & &\\ & \lambda_k & 1 & &\\ & & \ddots & \ddots &\\ & & & \lambda_k &1\\ & & & & \lambda_k\\ \end{bmatrix} \in \mathbb{C}^{N_k\times N_k} \end{equation} where $\lambda_k$ is an eigenvalue of multiplicity $N_k$ $(\sum\limits_{k=1}^p N_k = N)$

Answer 2

Yes, this is true. In the Jordan form of $M$, there will be precisely $r$ blocks associated to the eigenvalue $\lambda$. In each such block, there will be precisely one column (the first, by usual conventions) that contains $\lambda$ and only $\lambda$ and all other columns in the block will also contain a $1$.