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If $ A = BDB^{-1} $, $B \in Gl_n(K)$ and $ D = (d_{ij}) $ an upper triangular matrix with 1 on the diagonal line.

Show that A is unipotent, using the definition that a matrix A is unipotent if there is a $k \in N$ so that $(A - E_n)^k = 0$ where $E_n$ is the identity matrix.

For me, it sounds plausible that a triangular matrix with 1 on the diagonal is unipotent because it results in a nilpotent matrix if $E_n$ is subtracted. But I'm not sure how to show this for similar matrices.

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Try writing $A-E_n=BDB^{-1}-BE_nB^{-1}$ and use the fact that $D-E_n$ is nilpotent.

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  • $\begingroup$ Thanks, but could you please explain a bit further? I don't quite see why $BDB^{-1} - BE_nB^{-1}$ must be nilpotent just because $D - E_n$ is. How can we get rid of these these $B$ and $B^{-1}$? $\endgroup$ – moran Jan 18 '15 at 21:57
  • $\begingroup$ Factor $B$ out on the left and $B^{-1}$ on the right. Then compute powers of $B(D-E_n)B^{-1}$ explicitly in terms of powers of $D-E_n$. $\endgroup$ – Kevin Carlson Jan 18 '15 at 22:00
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A square matrix $M$ is a unipotent matrix if and only if its characteristic polynomial $P(t)$ is a power of $t − 1$. Equivalently, $M$ is unipotent if and only if all its eigenvalues are $1$. A matrix $A$ which is similar to $D$ as above has all eigenvalues equal to $1$, because similar matrices have the same eigenvalues. Hence $A$ is unipotent.

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