# Why is the similar of a triangular matrix unipotent

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.

Try writing $A-E_n=BDB^{-1}-BE_nB^{-1}$ and use the fact that $D-E_n$ is nilpotent.
• 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 $B^{-1}$? – moran Jan 18 '15 at 21:57
• 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 $D-E_n$. – Kevin Carlson Jan 18 '15 at 22:00
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.