Canonical form of linear operators- upper triangular equivalence

**davesface** · December 7th 2010, 10:00 PM

Show that all upper triangular matrices with diagonal entries $\alpha$ and arbitrary other elements are equivalent if $a_{12},a_{23},...,a_{n-1,n}$ are nonzero

**FernandoRevilla** · December 8th 2010, 01:53 AM

Consider

$A=\begin{bmatrix} \alpha & a_{12} & \ldots & a_{1n}\\ 0 &\alpha & \ldots & a_{2n} \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \alpha \end{bmatrix}\in \mathbb{R}^{n\times n}$

with $a_{12}\cdot a_{23}\cdot \ldots \cdot a_{n-1,n}\neq 0$ , then

$\lambda=\alpha$ is the only eigenvalue of $A$ (multiplicity $n$ ).

Besides,

$\dim \ker(A-\alpha I)=n-r(A-\alpha I)=n-(n-1)=1$ .

This means that the canonical form of Jordan $J$ for $A$ has only one block:

$J=\begin{bmatrix} \alpha & 1 & \ldots & 0\\ 0 &\alpha & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & \alpha \end{bmatrix}$

As a consequence, all matrices $A$ are equivalent to $J$ (even more, similar to $J$ ).

Regards.

Fernando Revilla

**FernandoRevilla** · December 8th 2010, 02:12 AM

I add the following to my previous post:

(i) If $\alpha\neq 0$ then, $r(A)=n$ and all matrices $A$ are equivalent to $I_n$ (by a well known theorem).

(ii) If $\alpha=0$ then, $r(A)=n-1$ and all matrices $A$ are equivalent to

$\begin{bmatrix}{I_{n-1}}&{0}\\{0}&{0}\end{bmatrix}$ (by a well known theorem).

So, we can avoid similarity and only use the concept of equivalence.

Regards.

Fernando Revilla

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