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I am studying Jordan normal form and still trying to piece it together. Got stuck on this question.

Given a matrix representation of $T: V \rightarrow V$ in basis $A = (v_1,..,v_n)$ by \begin{bmatrix} \lambda & 0 & 0 & \dots & 0 & 0 \\ 1 & \lambda & 0 & \dots & 0 & 0 \\ 0 & 1 & \lambda & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \lambda & 0 \\ 0 & 0 & 0 & \dots & 1 & \lambda \end{bmatrix}

In which basis $B$ has $M_{B}(T)$ Jordan normal form?

$M_{A}(T)$ has ones below the diagonal instead, so I am guessing there's an easy way to figure $B$ out.

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Hint: Use $B = (v_n,v_{n-1},\dots,v_1)$

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  • $\begingroup$ In general, reversing the order of your basis "flips the matrix upside down" $\endgroup$ – Ben Grossmann Jun 4 '18 at 23:15

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