11 - Jordan Blocks

Jordan Block

Let $J_n$ be a $(n\times n)-$matrix then

$$J_i:=\left(\begin{array}{cccc}0& 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & 0\end{array}\right)$$

Jordan Canonical Form for Nilpotent Operators corollary

Let $V$ be finite-dimensional and $T:V\to V$ be a linear transformation
Assume $m_{T}x)=(x-\lambda {)}^m$ then

There exists basis $\mathcal{B}$ such that ${}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ is block diagonal with blocks of form

$$J_i(\lambda ):=\lambda I_i+J_i=\left(\begin{array}{cccc}\lambda & 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & \lambda \end{array}\right)$$

Proof

As $m_T(x)=(x-\lambda {)}^m$ then $T-\lambda I$ is Nilpotent with minimal polynomial $x^m$

Applying Jordan Canonical Form for Nilpotent Operators then
So there exists basis of $\mathcal{B}$ such that ${}_{\mathcal{B}}[T-\lambda I{]}_{\mathcal{B}}$ is block diagonal with blocks $J_i$ hence

$${}_{\mathcal{B}}[T{]}_{\mathcal{B}}=\lambda I+{}_{\mathcal{B}}[T-\lambda I{]}_{\mathcal{B}}$$

is of the desired form