16 - Jordan Canonical Form Theorem

Jordan Canonical Form Theorem

Let $V$ be finite dimensional
Let $T:V\to V$ be a linear map with minimal polynomial

$$m_T(x)=(x-{\lambda }_1{)}^{m_1}\cdots (x-{\lambda }_r{)}^{m_r}$$

Then there exists basis $\mathcal{B}$ of $V$ such that ${}_{\mathcal{B}}[T{]}_{\mathcal{B}}$ is block diagonal
where each diagonal block is of the form

$$J_i({\lambda }_j)\text{ for one }1\le i\le m_j\text{ and }1\le j\le r$$

Notes

If $\mathbf{F}$ is an algebraically closed field then the minimal polynomial always will be in the form above

There could be several $J_i({\lambda }_j)$ for each pair $(i,j)$ or none
but there is at least one block for $i=m_j$ for $1\le j\le r$

For each $1\le j\le r$ then the number of Jordan Blocks for $1\le i\le m_j$
determines and is determined by the sequence of dimensions

$$\dim \mathrm{Ker}((T-{\lambda }_{j}I{)}^i)\text{ for }1\le i\le m_j$$

As the sequence of dimensions depends only upon $T$ then

$$\text{Jordan Form is unique up to the ordering of blocks}$$

Proof

By Primary Decomposition Theorem then

$$V=\mathrm{Ker}(T-{\lambda }_{1}I{)}^{m_1}\oplus \cdots \oplus \mathrm{Ker}(T-{\lambda }_{r}I{)}^{m_r}$$

Furthermore, $T$ restricted to the $j-$th summand has minimal polynomial $(x-{\lambda }_j{)}^{m_j}$
Hence apply Jordan Canonical Form for Nilpotent Operators