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The question asks me to prove that the matrix,

$$A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$

is neither unipotent nor nilpotent. However, can't I simply row reduce this to the identity matrix:

$$A=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$

which shows that it clearly is unipotent since $A^k=I$ for all $k \in \Bbb Z^+$?

Is there something wrong with the question or should I treat A as the first matrix rather than reducing it (I don't see how it would then not be unipotent considering the two matrices here are equivalent)?

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$A$ is unipotent, since $(A-I)^2= 0$. – Calvin Lin Jan 26 '13 at 22:25
Also, row reduction doesn't mean that you can always reduce a matrix to the identity matrix. For example, if you start with the 0 matrix, then you can only get the 0 matrix. – Calvin Lin Jan 26 '13 at 22:26
up vote 1 down vote accepted

HINT: As Calvin Lin pointed out in the comments, $(A-I)^2=0$, so $A$ is unipotent. To show that $A$ is not nilpotent, show by induction on $n$ that

$$A^n=\begin{bmatrix}1 & n\\0 & 1\end{bmatrix}\;.$$

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Or more easily, determinant of A is nonzero so it can not be nilpotent.

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