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Let $A,B,U \in SL_2(F)$. If $A$ and $B$ are upper-triangular matrices, then prove that $U^{-1}AU$ and $U^{-1}BU$ have common a eigenvector.

I know is that $[1,0]^t$ is a common eigenvector of $A$ and $B$. Any idea?

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    You changed $A$ and $B$ using $U$, so why not try and think of a way to change $[1,0]$ using $U$. – 2017-01-16
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    I tried it. We can multiply $Av=sv$ from the left by $U^{-1}$ but it is not possible multiply the equation from the right by $U$. – 2017-01-16
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    You were on the right lines, but multiplying the eigenvector equation directly won't give what you want. Another idea is to think about what $U^{-1}AU$ does: if $A$ is the matrix of a linear map with respect to the standard basis then $U^{-1}AU$ is the matrix of that linear map with respect to a different basis. What is $[1,0]^t$ in this new basis? – 2017-01-16
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    $U^{-1}AU(U^{-1}v)=s(U^{-1}v)$ – 2017-01-16
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    Got it :) So $U^{-1}[1,0]^t$ is the eigenvector – 2017-01-16

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