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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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Because that's not what the Jordan canonical form is. Are you asking whether such a matrix would be as simple to analyze as the Jordan form? Close, though if you look at the formulas for powers of the Jordan form you will see that the blocks have a very nice form, as do square roots and other matrices of interest. It's not clear that such would be the case with such a variant. –  Arturo Magidin Jul 15 '12 at 19:51
    
Ones and zeros. –  GEdgar Jul 15 '12 at 19:58
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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.

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