Jordan Blocks of the matrix with $a_{ij}\ne 0$ if $i = j + 1$ and $a_{ij}= 0$ if $i \geq j + 2$

Question

Let $A = (a_{jj})$ be a $n \times n$ complex matrix such that $a_{ij}\ne 0$ if $i = j + 1$ but $a_{ij}= 0$ if $i \geq j + 2$. Prove that A cannot have more than one Jordan block for any eigenvalue.

Answer 1

Let $\lambda$ be an eigenvalue of such a matrix $A$.

Recall that the geometric multiplicity of $\lambda$ as an eigenvalue of $A$ is $$ \DeclareMathOperator{gm}{gm}\gm_A(\lambda)=\dim\DeclareMathOperator{Null}{Null}\Null(\lambda\cdot I-A) $$ Also recall that $\gm_A(\lambda)$ is the number of Jordan blocks corresponding to the eigenvalue $\lambda$ in the Jordan canonical form $J$ of $A$.

Now, let $M$ be the $(1,n)$-submatrix of $\lambda\cdot I-A$. Note that $M$ is upper-triangular with nonzero diagonal entries (why?). Thus the $(1,n)$-minor of $\lambda\cdot I-A$ is nonzero. Since $\lambda\cdot I-A$ is singular, this implies that $\lambda\cdot I-A$ has rank $n-1$. The rank-nullity theorem then implires that
$$ \gm_A(\lambda)=\dim\Null(\lambda\cdot I-A)=n-\DeclareMathOperator{rank}{rank}\rank(\lambda\cdot I-A)=n-(n-1)=1 $$ Hence the number of Jordan blocks in the Jordan canonical form of $A$ associated to the eigenvalue $\lambda$ is one.

Answer 2

Hint. If $A$ has two or more Jordan blocks for some eigenvalue $\lambda$, then $A$ has at least two linearly independent eigenvectors corresponding to $\lambda$, meaning that the null space of $B=A-\lambda I$ is at least two-dimensional. What does it mean for the rank of $B$?