Lecture 7 - Jordan canonical form
In this sequence of videos, we will discuss the Jordan canonical form of a matrix/linear transformation.
Definition - Let \(U\) be a subspace of \(V\) and suppose \(T:V\to V\). Then \(U\) is called \(T\)-invariant if \(T(U)\subseteq U\).
Example - \(0\), \(V\), \(\mathrm{ker}(T)\), \(T(V)\), \(\mathrm{ker}(g(T))\), \(g(T)(V)\) where \(g\) is a polynomial, are all examples of invariant subspaces.
Key Lemma - \(V = \mathrm{ker}(g(T)) \oplus \mathrm{ker}(h(T))\) if \(f(T) = 0\) and \(f = gh\) with \(\gcd(g,h) = 1\).
Example - Suppose that \((T-2)(T-3) = 0\), then prove that \(T\) is diagonalizable.
Using the above lemma and induction, we can establish that
Theorem
\[V = \mathrm{ker}((T-\lambda_1)^{m_1})\oplus \mathrm{ker}((T-\lambda_2)^{m_2})\oplus\cdots \oplus \mathrm{ker}((T-\lambda_k)^{m_k}).\]The key is that each of the above subspaces is \(T\) invariant.
So the matrix of \(T\) will be a block diagonal matrix
\[\left(\begin{array}{cccc} A_1\\&A_2\\&&\ddots\\ &&&A_k\end{array}\right)\]where each \(A_i\) is the matrix of \(T\) restricted to \(\mathrm{ker}((T-\lambda_i)^{m_i})\)
Restricting to \(\mathrm{ker}((T-\lambda_i)^{m_i})\) and substituting \(T' = T-\lambda_i\), we have reduced to the case when \(T^m = 0\).
In order to understand the case when \(T:V\to V\) and \(T^m = 0\) we have cyclic subspaces.
Definition - Suppose that \(T^m\mathrm{v} = \mathrm{0}\) but \(T^{m-1}\mathrm{v} \neq \mathrm{0}\). Then \(U = \mathrm{span}(\mathrm{v},T\mathrm{v},\ldots,T^{m-1}\mathrm{v})\) is called a cyclic subspace of \(V\). The vector \(\mathrm{v}\) is called a cyclic vector. Notice that every element of \(U\) can be written as \(\mathrm{u} = p(T) \mathrm{v}\) for some polynomial \(p\). If \(U\) is a cyclic subspace with cyclic vector \(\mathrm{v}\) we write \(U = C(\mathrm{v})\).
First, we define cyclic subspaces and prove that \(\mathrm{v},T\mathrm{v},\ldots, T^{m-1}\mathrm{v}\) is a basis for \(C(\mathrm{v})\).
Then we prove that if \(T:V\to V\) and \(T^m = 0\), and \(V\) is finite-dimensional, then \(V\) is the direct sum of cyclic subspaces.
The proof is by induction.
Finally, we (briefly) mention the payout from all this work.