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In the Jordan form of square matrix $A \longrightarrow T^{-1}AT = J$, $J$ needs to be upper bidiagonal; but should the upper diagonal be restricted to ones?.

The equations $Av_i = v_{i-1} + \lambda_iv_i $, where $v_i$ are the columns of $T$, result from the Jordan form and they establish the linear independence of $T$'s columns. Why cant we have $Av_i = 2v_{i-1} + \lambda_iv_i $ with upper diagonal being 2 or just any number.

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    Ones and zeros.2012-07-15

2 Answers 2

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Upper diagonal does not have to be with $1$s, it just happens to be a multiplicative unit and also the world's favorite nonzero number. If you find the Jordan form of $A/2$ and multiply the result by $2$, you'll get a matrix with $2$ instead of $1$ in the upper diagonal, as you wanted. This works with any nonzero number too.

The logic of having $1$ above the diagonal is that they form the shift operator, which is the simplest (and also world's favorite) non-normal operator. If $A$ is normal, there is nothing above the diagonal in the Jordan form; otherwise it has some amount of shift going on.

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Actually, sometimes it is useful to look at "pseudo"-Jordan forms where instead of $1$s, you have some fixed real number $r$ in the same places. It's an easy exercise that after getting it into Jordan canonical form, there's another basis change (depending on $r$) you can do to get it into such a form.

In particular, it is useful to look at $r \to 0$. See for example, chapter 22 of V. I. Arnold's Ordinary Differential Equations.