# The impossibility of decomposing the lower shift operator

I am wondering about the quickest way to prove the following from scratch: the lower shift is not similar to a direct sum of other linear transformations.

Here are the definitions of the terms used in the question. The lower shift is a mapping from $R^n$ to $R^n$ which maps $(x_1, x_2, \ldots, x_{n-1}, x_n)$ to $(0, x_1, \ldots, x_{n-1})$. Given two linear mappings $T_1: U \rightarrow U, ~T_2: V \rightarrow V$ their direct sum is the map from $U \times V~$ to $~U \times V$ which maps $(u,v)$ to $(T_1 u, T_2 v)$.

My current proof of this appeals to the Cayley-Hamilton theorem.

Thank you!

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## 2 Answers

Tensor by $\mathbb{C}$. Then the direct sum of two linear transformations has at least two eigenvectors, but the lower shift has only one. (If you don't want to tensor by $\mathbb{C}$, the corresponding statement over $\mathbb{R}$ is that the direct sum of two linear transformations has at least two one- or two-dimensional invariant subspaces, but the lower shift has only one.)

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Thanks, that is very clean and quick. –  angela o. Oct 10 '10 at 23:48

The shift is nilpotent. If it were nontrivially a direct sum, then each of the summands would be nilpotent on spaces of positive dimension. The kernel of a nilpotent on a nontrivial space is nontrivial, and the kernel of a direct sum is the direct sum of the kernels, so this would imply that the kernel of the shift has dimension at least 2. But its kernel is 1-dimensional.

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