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This was a problem from a homework several weeks ago. I've got a test coming up, and I'm still not sure I understand it. The professor mentioned it having something to do with the $(n-1)\times (n-1)$ submatrix having full rank, but I'm still unsure about it.

Let $H \in \mathbb{C}^{n \times n}$ be an irreducible Hessenberg matrix, i.e., $$ H=\begin{bmatrix} h_ {11} & h_ {12}& \cdots & h_ {1,n-1}& h_ {1n}\\ h_ {21} & h_ {22 }& \cdots & h_ {2,n-1} & h_ {2n}\\ 0 & h_ {32} & \cdots & h_ {3,n-1} & h_ {3n}\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0& \cdots & h_ {n,n-1}&h_ {nn} \end{bmatrix}$$ with $h_{i+1,i} \ne 0$ for $ i=1:n-1$. Prove that the eigenvalues of $H$ are geometrically simple; that is, $d(\lambda)= 1, \forall \lambda \in \sigma(H)$.

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If $Hv=tv$, then $(H-tI)v=0$. In this matrix the block obtained by erasing the first row and the last column is diagonal with nonzero diagonal. So the rank of $H-tI$ is $n-1$, which tells us that its kernel has dimension $1$. This works for any eigenvalue $t$, so all eigenvalues have geometric multiplicity one.

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