Editing Nilpotent matrix (section)

==Classification==
Consider the <math>n \times n</math> (upper) [[shift matrix]]:
:<math>S = \begin{bmatrix} 
   0 & 1 & 0 & \ldots & 0  \\
   0 & 0 & 1 & \ldots & 0  \\
   \vdots & \vdots & \vdots & \ddots & \vdots \\
   0 & 0 & 0 & \ldots & 1  \\
   0 & 0 & 0 & \ldots & 0
\end{bmatrix}.</math>
This matrix has 1s along the [[superdiagonal]] and 0s everywhere else.  As a linear transformation, the shift matrix "shifts" the components of a vector one position to the left, with a zero appearing in the last position:
:<math>S(x_1,x_2,\ldots,x_n) = (x_2,\ldots,x_n,0).</math><ref>{{harvtxt|Beauregard|Fraleigh|1973|p=312}}</ref>
This matrix is nilpotent with degree <math>n</math>, and is the [[Canonical form|canonical]] nilpotent matrix.

Specifically, if <math>N</math> is any nilpotent matrix, then <math>N</math> is [[matrix similarity|similar]] to a [[block diagonal matrix]] of the form
:<math> \begin{bmatrix} 
   S_1 & 0 & \ldots & 0 \\ 
   0 & S_2 & \ldots & 0 \\
   \vdots & \vdots & \ddots & \vdots \\
   0 & 0 & \ldots & S_r 
\end{bmatrix} </math>
where each of the blocks <math>S_1,S_2,\ldots,S_r</math> is a shift matrix (possibly of different sizes).  This form is a special case of the [[Jordan canonical form]] for matrices.<ref>{{harvtxt|Beauregard|Fraleigh|1973|pp=312,313}}</ref>

For example, any nonzero 2&nbsp;&times;&nbsp;2 nilpotent matrix is similar to the matrix
:<math> \begin{bmatrix} 
   0 & 1 \\
   0 & 0
\end{bmatrix}. </math>
That is, if <math>N</math> is any nonzero 2&nbsp;&times;&nbsp;2 nilpotent matrix, then there exists a basis '''b'''<sub>1</sub>,&nbsp;'''b'''<sub>2</sub> such that ''N'''''b'''<sub>1</sub>&nbsp;=&nbsp;0 and ''N'''''b'''<sub>2</sub>&nbsp;=&nbsp;'''b'''<sub>1</sub>.

This [[classification theorem]] holds for matrices over any [[field (mathematics)|field]].  (It is not necessary for the field to be algebraically closed.)