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\begin{document}
\centerline{Math 322 lecture for Wednesday, Week 4}
\bigskip

{\bf Review of diagonalization.} For a diagonal
matrix~$D=\diag(\lambda_1,\dots,\lambda_n)$, we have
  \[
De_i=\lambda_ie_i
  \]
  for each standard basis vector~$e_i$.  If~$A\in M_n(F)$ is not diagonal, we
  look for linearly independent vectors that behave like the~$e_i$ above:
  \[
    Av_i=\lambda_i v_i.
  \]
  If we can find~$n$ of these vectors, then changing to the basis~$\left\{
  v_1,\dots,v_n \right\}$, these~$v_i$ are transformed to the standard basis
  vectors in the new coordinates, and~$A$ is diagonalized.

  Therefore, we look for vectors~$v\neq0$ such that
  \[
Av=\lambda v
  \]
  for some~$\lambda\in F$.  We have
  \[
    Av=\lambda v\quad\Leftrightarrow\quad (A-\lambda I_n)v=0
    \quad\Leftrightarrow\quad v\in\ker(A-\lambda I_n).
  \]
  The kernel is nonzero if and only if~$\det(A-\lambda I_n)=0$.  So to find
  suitable~$\lambda$, the {\em eigenvalues}, we consider the {\em characteristic
  polynomial}
  \[
    p(x)=\det(A-xI_n)=\prod_{j=1}^n(\lambda_j-x)=\prod_{j=1}^{\ell}(\mu_j-x)^{k_j}.
  \]
  In the expression on the far right, repeated eigenvalues are grouped together
  (so each~$\mu_j$ is equal to some~$\lambda_t$).  The {\em algebraic
  multiplicity} of the eigenvalue~$\mu_j$ is~$k_j$.  The eigenvectors
  corresponding to~$\mu_j$ form a subspace of~$F^n$ called the {\em eigenspace}
  for~$\mu_j$:
  \[
    E_{\mu_j}:=\ker(A-\mu_jI_n).
  \]
  The dimension of~$E_{\mu_j}$ is the {\em geometric multiplicity} of~$\mu_j$.
  We always have that the geometric multiplicity is at most the algebraic
  multiplicity:
  \[
    \dim E_{\mu_j}\leq k_j.
  \]
  The matrix~$A$ is diagonalizable if and only if there is a basis consisting of
  eigenvectors, and that happens exactly when the geometric multiplicity of each
  eigenvalue equals it algebraic multiplicity.  If that is not the case, we can
  still choose bases for each eigenspace, but we are then left with the task of
  completing this set to a full basis for~$F^n$.  By choosing correctly, we can
  assure that~$A$ has a nice form.
  \bigskip

\centerline{\sc jordan form}

Let~$\lambda\in F$. A {\em $k\times k$ Jordan block} for~$\lambda$ is
a~$k\times k$ matrix with~$\lambda$ appearing along the diagonal and~$1$s
appearing on the superdiagonal:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle
    J_k(\lambda):=
    \left(\begin{array}{cccccc}
	\lambda&1&&&&\\
	&\lambda&1&&&\\
	&&\lambda&1&&\\
	&&&\ddots&&\\
        &&&&&1\\
       &&&&&\lambda
 \end{array} \right).$};
 \node at (-0.2,-0.6) {\huge 0};
 \node at (1.9,1) {\huge 0};
  \end{tikzpicture}
\end{center}
  For example,
  \[
    J_4(2)=
    \left(\begin{array}{cccc}
	2&1&0&0\\
	0&2&1&0\\
	0&0&2&1\\
	0&0&0&2
    \end{array} \right).
  \]
  A {\em Jordan matrix} is a square block-diagonal matrix with Jordan matrices
  along the diagonal:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle
    J:=
    \left(\begin{array}{cccccc}
	J_{k_1}(\lambda_1)&&&&&\\
	  &J_{k_2}(\lambda_2)&&&&\\
	  &&J_{k_3}(\lambda_3)&&&\\
	&&&\ddots&&\\
        &&&&&\\
	&&&&&J_{k_{\ell}}(\lambda_{\ell}).
 \end{array} \right).$};
 \node at (-1.2,-0.6) {\huge 0};
 \node at (2.6,1) {\huge 0};
  \end{tikzpicture}
\end{center}
For example, the following is a Jordan matrix:
\[
  \left(\begin{array}{cccccccc}
      2&0&0&0&0&0&0&0\\
      0&2&0&0&0&0&0&0\\
      0&0&4&1&0&0&0&0\\
      0&0&0&4&1&0&0&0\\
      0&0&0&0&4&0&0&0\\
      0&0&0&0&0&i&1&0\\
      0&0&0&0&0&0&i&0\\
      0&0&0&0&0&0&0&2+3i\\
  \end{array} \right)
\]
with Jordan blocks~$J_1(2), J_1(2),J_3(4),J_2(i)$ and~$J_1(2+3i)$.

A diagonal matrix is a Jordan matrix whose Jordan blocks are all~$1\times1$.
\medskip

\textbf{Theorem.} Let~$A\in M_n(\C)$.  Then there exists an invertible
matrix~$P\in M_n(\C)$ such that~$P^{-1}AP=J$ where~$J$ is a Jordan matrix.  The
matrix~$J$ is called the {\em Jordan form for~$A$}.  It is unique up to a
permutation of the Jordan blocks.  The diagonal entries of~$J$ are exactly the
eigenvalues of~$A$ repeated according to their algebraic multiplicities
  (the number of times the eigenvalue appears in a factorization of the
characteristic polynomial of~$A$ over~$\C$).  The number of blocks having a particular
eigenvalue~$\lambda$ along the diagonal is the geometric multiplicity
of~$\lambda$ (i.e.,~$\dim(A-\lambda I_n)$).
\medskip

{\bf Example.} A matrix is diagonalizable if and only if each of its Jordan
blocks is~$1\times 1$.  For example, we know
\[
  A=
  \left(\begin{array}{cc}
      1&1\\
      0&1
  \end{array} \right)
\]
is not diagonalizable
since it is already in Jordan form and it's not diagonal.  The matrix~$A$ has one
eigenvalue,~$1$, of multiplicity~$2$, but the dimension of the eigenspace
for~$1$ is $1$-dimensional:
\[
  \ker(A-1\cdot I_2)=\ker
  \left(\begin{array}{cc}
      0&1\\
      0&0
  \end{array} \right)=\left\{ (x,0) :x\in F\right\},
\]
which has basis~$\left\{ (1,0) \right\}$.  As claimed the number of Jordan
blocks for~$1$ is the geometric multiplicity of~$1$.
\medskip

{\bf Jordan form over the reals.} 
Now suppose that~$A\in M_n(\R)$.  Then it turns out that we can conjugate~$A$ via a real matrix it to a real matrix
that is almost as nice as the Jordan form over~$\C$.  Since~$A$ is defined over
the reals, its nonreal eigenvalues appear in conjugate pairs, and it turns out
that each~$k\times k$ Jordan block for~$\lambda=a+bi$ has a corresponding~$k\times k$
Jordan block for~$\bar{\lambda}=a-bi$ of the same dimension.  We can combine these
blocks and change basis to get a corresponding~$2k\times 2k$ block matrix
with~$2\times 2$ blocks of the form
\[
  \left(\begin{array}{cc}
      a&-b\\b&a
  \end{array} \right)
\]
along the diagonal, and the~$2\times 2$ identity matrix~$I_2$ appearing along the super
diagonal.  For instance, the Jordan matrix
\[
  \left(\begin{array}{cccccc}
      \lambda&1&0&0&0&0\\
      0&\lambda&1&0&0&0\\
      0&0&\lambda&0&0&0\\
      0&0&0&\bar{\lambda}&1&0\\
      0&0&0&0&\bar{\lambda}&1\\
      0&0&0&0&0&\bar{\lambda}\\
  \end{array} \right)
\]
where~$\lambda=a+bi$ can be conjugated to the form
\[
  \left(\begin{array}{cccccc}
      a&-b&1&0&0&0\\
      b&a&0&1&0&0\\
      0&0&a&-b&1&0\\
      0&0&b&a&0&1\\
      0&0&0&0&a&-b\\
      0&0&0&0&b&a\\
  \end{array} \right)
\]
If~$A\in M_n(\R)$, there exists an invertible~$P\in M_n(\R)$ such
that~$P^{-1}AP=J$ where~$J$ consists of Jordan blocks---the usual ones for real
eigenvalues, and these modified block matrices for conjugate pairs of
complex eigenvalues.  The form is unique up to permutation of the blocks and swaps
\[
  \left(\begin{array}{cc}
      a&-b\\b&a
  \end{array} \right)
  \quad
  \longleftrightarrow
  \quad
  \left(\begin{array}{cc}
      a&b\\-b&a
  \end{array} \right).
\]
We will call it the {\em real Jordan form} for~$A$.
Here is a typical real Jordan form for a real matrix:
\[
  \left(\begin{array}{cccccccccc}
      4&0&0&0&0&0&0&0&0&0\\
      0&4&1&0&0&0&0&0&0&0\\
      0&0&4&1&0&0&0&0&0&0\\
      0&0&0&4&0&0&0&0&0&0\\
      0&0&0&0&3&-2&0&0&0&0\\
      0&0&0&0&2&3&0&0&0&0\\
      0&0&0&0&0&0&3&-2&1&0\\
      0&0&0&0&0&0&2&3&0&1\\
      0&0&0&0&0&0&0&0&3&-2\\
      0&0&0&0&0&0&0&0&2&3\\
  \end{array} \right)
\]
There are two Jordan blocks for~$4$: one is~$1\times1$ and one is~$3\times3$.
The other eigenvalues for this matrix are~$3+2i$ and~$3-2i$, each of which
appears with multiplicity~$3$. Notice there are two real Jordan blocks for the
pair~$3\pm2i$, one is~$2\times 2$ and the other is~$4\times 4$.
\end{document}
