A = B+ I, B^2 = 0

Let {A = \begin{pmatrix} a + 1& -a\\ a & -a + 1\\ \end{pmatrix}}. What can we say about {A}?

Remark: {A} has some interesting properties: {A = B + I} where {B^2 = O}.

  1. {A^n = nB+I, n \in \mathbb{Z}}.
    Indeed, we can prove it by induction.
    {A^2 = (B+I)(B+I) = B^2 + 2B + I = 2B+I},  {(B+I)(-B+I) = I \Rightarrow A^{-1} = -B + I, A^{-2} = (-B + I)(-B+I) = -2B+I\dots} Suppose that {A^k = kB+I}, we have {A^{k+1} = (kB+ I)(B+I) = kB^2 + (k+1)B + I = (k+1)B + I}. Moreover, suppose that {A^{-k} = -B+I}, we have {A^{-k-1} = (-kB+I)(-B+I) = -(k+1)B + I}.
  2. {A^{-1} = - B+I \Leftrightarrow B + I = - A^{-1} + 2I \Leftrightarrow A = -A^{-1} + 2I \Leftrightarrow A + A^{-1} = 2I}.
Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s