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We know the Jordan block $$J = \begin{bmatrix} \lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1\\ & & & \lambda \end{bmatrix} $$

Prove that there exists an invertible $\bf{S}$ such that $$\bf{SJS^{-1}} = \begin{bmatrix} \lambda & \varepsilon & & \\ & \lambda & \ddots & \\ & & \ddots & \varepsilon\\ & & & \lambda \end{bmatrix} $$ with any nonzero $\varepsilon$.

I have no idea how to construct the $\bf{S}$, or how to 'diagonalize' the Jordan block. Could anyone give me some hints? Thanks in advance!

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    $\begingroup$ As always: Start with a 2×2 matrix. Then try 3×3. You should see a pattern. $\endgroup$ – P. Siehr Jan 9 '18 at 9:39
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    $\begingroup$ The Jordan block can't be diagonalised (that's why we talk about Jordan blocks in the first place). $\endgroup$ – user228113 Jan 9 '18 at 10:36
  • $\begingroup$ @G.Sassatelli Yep, I made a mistake. I would edit the question. $\endgroup$ – stander Qiu Jan 9 '18 at 12:07
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Take$$S=\begin{bmatrix}1&0&0&\cdots&0\\0&\varepsilon^{-1}&0&\cdots&0\\0&0&\varepsilon^{-2}&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\varepsilon^{-(n-1)}\end{bmatrix}$$(where $n$ is such that $J$ is a $n\times n$ matrix). Then$$SJS^{-1}=\begin{bmatrix}\lambda&\varepsilon&0&\cdots&0\\0&\lambda&\varepsilon&\cdots&0\\0&0&\lambda&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda\end{bmatrix}.$$

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    $\begingroup$ If OP asks "give me some hints", I think it's better not to answer with a complete solution. "Try with a diagonal $S$" would have been a better answer in my view. $\endgroup$ – Federico Poloni Jan 9 '18 at 15:39
  • $\begingroup$ @FedericoPoloni If I had thought about that suggestion, perhaps that I would have provided it. But what happens quite often when I provide just a hint is what happend here today; someone else provides a complete answer after me and my answer is ignored. $\endgroup$ – José Carlos Santos Jan 9 '18 at 19:21
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By Jordan normal form theorem, if $A\in k^{n\times n}$ is triangulable and $B\in k^{n\times n}$, then $A$ and $B$ are similar if and only if $\dim\ker (A-\lambda I)^m=\dim\ker (B-\lambda I)^m$ for all $m\in\Bbb N$ and for all $\lambda\in k$.

This is easily applied to this case, because $J_\varepsilon-\lambda I=\varepsilon(J-\lambda I)$, while $J_\varepsilon-\mu I$ and $J-\mu I$ are both invertible when $\mu\ne \lambda$.

The same lemma and some calculations prove that an upper-triangular matrix is similar to $J$ if and only if, for all $i$, $a_{i,i}=\lambda$ and $a_{i,i+1}\ne 0$.

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