# Linear Algebra : Jordan Canonical form (Jordan blocks and the Super-Diagonal)

In terms of Jordan Canonical Form, and more specifically about Jordan Blocks. When there is a definition about Jordan Blocks they say the eigenvalues go on the principle diagonal and the diagonal above it(usually called the super-diagonal) contains the number 1!

I want to understand why there are 1's in the superdiagonal. I have looked all over and no body tries to explain why there is this mysterious 1's in the super-diagonal.

I hope that someone here on this forum can explain this.

Here is a link to PDF: http://ckottke.ncf.edu/docs/jordan.pdf

• "I have looked all over": Have you looked at a proof that every (complex) matrix has a Jordan form? – David C. Ullrich Dec 22 '18 at 16:03
• Hi David, i have seen things about complex matrix has Jordan form, but when they discuss it they assume the Ones on the Super Diagonal as though this is a fundamental axiom without need of explanation. – Palu Dec 22 '18 at 19:57
• I just added a link to a pdf article I want to illustrate, how they explain things, where they just assume it to be a fact. See above. – Palu Dec 22 '18 at 20:00

If you have a non-zero vector $$x$$ such that $$(A-\lambda I)^{n-1}x \ne 0$$ and $$(A-\lambda I)^{n}x=0$$, then $$\{x,(A-\lambda I)x,\cdots,(A-\lambda I)^{n-1}x\}$$ is a linearly independent set of vectors. Let \begin{align} v_1 & = (A-\lambda I)^{n-1}x,\\v_2 & =(A-\lambda I)^{n-2}x,\\&\cdots\\v_{n-1} & =(A-\lambda I)^{1}x\end{align}
Then $$A-\lambda I$$ has the following matrix representation with respect to this basis: $$\left[\begin{array}{cccc}0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \end{array}\right]$$ And $$A-\lambda I$$ has the representation $$\lambda I$$ plus the above, which is a Jordan block. $$\left[\begin{array}{cccc}\lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \lambda & \cdots & 1 \\ 0 & 0 & 0 & \cdots & \lambda \end{array}\right]$$ So the answer to your question is that you can always choose the basis so that you get $$1$$'s on the diagonal above the main diagonal for a Jordan block.
• By a change of basis, the k-th basis element becomes $e_k$. Jordan form is achieved by a change of basis – Disintegrating By Parts Dec 24 '18 at 2:49
• So when you use $e_k$ i guess you mean they are the standard basis of $R^n$. For example (1,0,0) , (0,1,0) and (0,0,1) for $R^3$. – Palu Dec 24 '18 at 16:07