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
 A: 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.
