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Let $K$ be some field and $V$ an $n$-dimensional vector space over $K$. Let $\alpha, \beta \in GL(V)$ be two linear maps, and choose some basis $\{ u_1, u_2, \ldots, u_n \}$ such that $\alpha$ has the representation matrix $$ \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & & & \ddots \\ a & 0 & 0 & \cdots & 1 \end{pmatrix} $$ and $\beta$ has the matrix $$ \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & & & \ddots \\ b & 0 & 0 & \cdots & 1 \end{pmatrix} $$ for some $a,b \in K$.

How to prove there exists some linear map $\gamma : V \to V$ such that $$ \alpha = \gamma^{-1} \beta \gamma $$ i.e. the maps are conjugate?

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Hint: Try choosing $\gamma$ to be a diagonal matrix. (Also, you have to assume $a$ and $b$ are nonzero (or else that they are both zero) for the result to be true.)

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  • $\begingroup$ Yes. $$ \begin{pmatrix} b & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & & \ddots \\ 0 & 0 & \cdots & a \end{pmatrix} $$ works! $\endgroup$
    – StefanH
    Nov 15, 2015 at 21:55

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