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\begin{document}
\centerline{Math 201 lecture for Friday, Week 10}
\bigskip

\centerline{\textbf{Algebraic and geometric multiplicity. Jordan form.}}  
\bigskip
 
\emph{When does a transformation fail to be diagonalizable?} We now introduce
a sequence of ideas that will allow us to answer this question.
\medskip

\textbf{Example.} Earlier, we considered the linear transformation~$\R^2\to\R^2$
given by the matrix
\[
A=
\left(\begin{array}{cc}
    0&-1\\
    1&0
\end{array} \right).
\]
Geometrically, it rotates the plane counterclockwise by~$90^{\circ}$ and, hence,
has no eigenvectors: an eigenvector would not rotate---it would just be scaled.
The characteristic polynomial of~$A$ is
\[
  p_A(x)=\det
\left(\begin{array}{cc}
    -x&-1\\
    1&-x
\end{array} \right)
=x^2+1.
\]
The equation~$x^2+1=0$ has no solutions over~$\R$, and hence, the transformation
has no eigenvalues.

Now consider the linear transformation~$f\colon\C^2\to\C^2$ given by the same
matrix~$A$.  Over~$\C$ we can solve~$x^2+1=0$ to find two
eigenvalues,~$\pm i$.  Each of these will have at least one eigenvector, and
eigenvectors for distinct eigenvalues are linearly independent.
Since~$\C^2$ has dimension~$2$, that means we will get a basis of eigenvectors.
Let's compute a basis for the eigenspace for~$i$:
\[
  A-iI_2=
  \left(\begin{array}{cc}
      -i&-1\\
      1&-i
  \end{array} \right)
  \quad\xrightarrow{r_1\leftrightarrow r_2}
  \left(\begin{array}{cc}
      1&-i\\
      -i&-1
  \end{array} \right)
  \quad\xrightarrow{r_2\leftrightarrow r_2+ir_1}
  \left(\begin{array}{cc}
      1&-i\\
      0&0
  \end{array} \right).
\]
So the kernel of~$A-iI_2$ is~$\left\{ (iy,y):y\in\C \right\}$, which has
basis~$\left\{ (i,1)\right\}$. Similarly, the eigenspace for~$-i$ has
basis~$\left\{ (-i,1) \right\}$.  Check:
\begin{align*}
  A
  \left(\begin{array}{c}
      i\\1
  \end{array} \right)
  &=
\left(\begin{array}{cc}
    0&-1\\
    1&0
\end{array} \right)
  \left(\begin{array}{c}
      i\\1
  \end{array} \right)
  =
  \left(\begin{array}{c}
      -1\\i
  \end{array} \right)
  =
  i
  \left(\begin{array}{c}
      i\\1
  \end{array} \right)\\[10pt]
  A
  \left(\begin{array}{c}
     -i\\1
  \end{array} \right)
  &=
\left(\begin{array}{cc}
    0&-1\\
    1&0
\end{array} \right)
  \left(\begin{array}{c}
      -i\\1
  \end{array} \right)
  =
  \left(\begin{array}{c}
      -1\\-i
  \end{array} \right)
  =
  -i
  \left(\begin{array}{c}
      -i\\1
  \end{array} \right).
\end{align*}
Letting
\[
P=
\left(\begin{array}{cc}
    i&-i\\
    1&1
\end{array} \right),
\]
we get
\[
  P^{-1}AP=
  \left(\begin{array}{cc}
      i&0\\
      0&-i
  \end{array} \right).
\]
This example illustrates one obstacle to diagonalization: the characteristic
polynomial may not have enough roots in the field~$F$.
\medskip

\textbf{Definition.} A polynomial~$p\in F[x]$ \emph{splits over~$F$} if there
exist~$c,\lambda_1,\dots,\lambda_n\in F$ such that
\[
  p(x)=c(x_1-\lambda_1)\cdots(x-\lambda_n).
\]
Equivalently,~$p(x)$ had~$n$ roots (zeros),~$\lambda_1,\dots,\lambda_n$, in~$F$.
These~$\lambda_i$ need not be distinct. 
\medskip

\textbf{Remark.} Let $F$ be any field, and let $p(x)$ be a polynomial whose
coefficients are in~$F$, i.e., $p(x)\in F[x]$.  It turns out that there exists a
field~$F\subseteq K$ such that~$p(x)$ splits over~$K$.
\medskip

\textbf{Example.} The polynomial~$p(x)=x^2+1$ splits over~$\C$ but not
over~$\R$.
\medskip

A useful fact from algebra:
\medskip

\textbf{Theorem.} (Fundamental theorem of algebra) Every~$p\in\C[x]$ splits
over~$\C$.
\medskip

\textbf{Proposition.} Let~$V$ be a vector space over~$F$ with~$\dim V=n$, and let~$f\colon V\to V$
be a linear transformation.  If~$f$ is diagonalizable, then
its characteristic polynomial splits over~$F$.

\begin{proof}
  Let~$D=\diag(\lambda_1,\dots,\lambda_n)$ be a diagonal matrix
  representing~$f$.  Then the characteristic polynomial for~$f$ (which, as we saw
    earlier, in the last lecture, does not depend on the choice of matrix
  representative) is
  \[
   p_{f}(x)=p_{D}(x)= \det(D-xI_n)=(\lambda_1-x)\cdots(\lambda_n-x)=(-1)^n(x_1-\lambda_1)\cdots(x-\lambda_n).
  \]
\end{proof}

The converse of this proposition is not true:
\medskip

\textbf{Example.} Let
\[
A=
\left(\begin{array}{cc}
    1&1\\
    0&1
\end{array} \right).
\]
The characteristic polynomial of~$A$ is
\begin{align*}
  p_A(x)&= \det(A-x I_2)\\
  &= 
  \det
  \left(\begin{array}{cc}
      1-x&1\\
      0&1-x
  \end{array} \right)\\[5pt]
  =(x-1)^2.
\end{align*}
Thus, the characteristic polynomial splits over any field~$F$. There is one
eigenvalue,~$1$, which occurs with algebraic multiplicity~$2$ (the precise definition of
\emph{algebraic multiplicity} appears below).  Let's proceed
with the algorithm for diagonalization by computing a basis for the eigenspace
for~$1$, i.e., for~$\ker(A-I_2)$:
\[
\left(\begin{array}{cc}
    1&1\\
    0&1
\end{array} \right)
-
\left(\begin{array}{cc}
    1&0\\0&1
\end{array} \right)
=
\left(\begin{array}{cc}
    0&1\\
    0&0
\end{array} \right).
\]
Therefore,~$\ker(A-I_2)=\left\{ (x,0): x\in F \right\}$.  A basis is~$\left\{
(1,0) \right\}$.  Thus, there is no basis for~$F^2$ consisting of eigenvectors:
our theory says any eigenvector would have to have eigenvalue~$1$, and the space
of eigenvectors with eigenvalue~$1$ is only one-dimensional!
\medskip

\textbf{Definition.} Let~$\dim V<\infty$. The \emph{algebraic multiplicity} of
an eigenvalue~$\lambda\in F$ for a linear transformation~$f\colon V\to V$ (or for any
matrix representing~$f$) is the largest number~$m$ such
that~$p_f(x)=(x-\lambda)^{m}q(x)$ for some polynomial~$q(x)\in F[x]$.

The \emph{geometric multiplicity} of~$\lambda$ is the dimension of the
eigenspace~$E_{\lambda}(f)$ for~$\lambda$:
\[
\dim E_{\lambda}(f) = \dim\ker(f-\lambda\,\mathrm{id}_V).
\]
So if~$A$ is a matrix representing~$f$, then the geometric multiplicity
of~$\lambda\in F$ is
\[
  \dim E_{\lambda}(A) = \dim\ker(A-\lambda\,I_n).
\]
\medskip

\textbf{Remark.} To rephrase something we already know: \textbf{$A\in M_{n\times
n}(F)$ is diagonalizable if and only if the sum of its geometric multiplicities
is~$n$.}  That's because this is the only case in which we have enough linearly independent
eigenvectors to form a basis of eigenvectors.
\medskip

\textbf{Proposition.} Let~$\dim V<\infty$, and let~$\lambda$ be an eigenvalue of a linear transformation
~$f\colon V\to V$.  Then the geometric multiplicity of~$\lambda$ is at most the
algebraic multiplicity of~$\lambda$.

\begin{proof}
  Let~$v_1,\dots,v_k$ be a basis
  for~$\ker(f-\lambda\,\mathrm{id}_V)$, and extend it to a basis~$v_1,\dots,v_n$
  for all of~$V$.  We have~$f(v_i)=\lambda v_{i}$ for~$1=1,\ldots,k$.  So with
  respect to our chosen basis, the matrix
  representing~$f$ has the form
  \[
  A := 
  \left(\begin{array}{cc}
    \lambda I_k&B\\
    0&C
  \end{array} \right),
  \]
  where~$B$ and~$C$ are~$(n-k)\times(n-k)$ matrices.  So the characteristic
  polynomial for~$f$ is
  \begin{align*}
    p_f(x)
    &= 
    \det
    \left(\begin{array}{cc}
        (\lambda-x)I_k&B\\
        0&C-xI_{n-k}
    \end{array} \right)\\[10pt]
    &= \det((\lambda-x)I_k)\det(C-xI_{n-k})\\[10pt]
    &= (\lambda-x)^k\det(C-xI_{n-k})\\[10pt]
    &= (\lambda-x)^kq(x),
  \end{align*}
  for some polynomial~$q(x)$.  (To see the second equality, above, expand the
    determinant in line~$1$ along the first \emph{column}---there will only be one term, which
    will be~$\lambda-x$ times a smaller determinant.  Expand that determinant along its first
  column.  Repeat~$k$ times, each time picking up a factor of~$\lambda-x$.)  This shows that the algebraic multiplicity
  of~$\lambda$ is at least~$k$, the geometric multiplicity of~$\lambda$.
\end{proof}

\textbf{Corollary.} Let~$A\in M_{n\times n}(F)$.  Then~$A$ is diagonalizable if
and only if its characteristic polynomial splits over~$F$ and the geometric
multiplicity and algebraic multiplicity of each eigenvalue are equal.
\begin{proof}
  Suppose that~$A$ is diagonalizable.  We saw earlier in this lecture that 
  the characteristic polynomial for~$A$ then splits over~$F$.  So we
  can write
  \[
    p_{A}(x)=(-1)^{n}\prod_{i-1}^{k}(x-\lambda_{i})^{m_{i}}  
  \]
  where~$\lambda_{1},\ldots,\lambda_{k}$ are the distinct eigenvalues of~$A$.
  The degree of~$p_{A}(x)$ is~$n$ (exercise!), from which it follows
  that~$\sum_{i=1}^{k} m_{i}=n$, i.e, the sum of the algebraic multiplicities
  is~$n$.  

  Let $g_{i}$ be the geometric multiplicity of eigenvalue~$\lambda_{i}$.
  Since~$A$ is diagonalizable, we know that the sum of its geometric
  multiplicities is also~$n$.  

  Therefore, we have~$n=\sum_{i=1}^{k} g_i=\sum_{i=1}^{k} m_i$, and by the
  Proposition, $g_{i}\leq m_{i}$ for all~$i$.  Since the $m_i$ are nonnegative,
  it follows that~$m_{i}=g_{i}$ for all~$i$.

  Conversely, suppose that~$p_{A}(x)$ splits and that the algebraic and
  geometric multiplicities of each eigenvalue are equal. Factor~$p_{A}(x)$
  as above and using same notation for algebraic and geometric
  multiplicities.   As before, since the degree of~$p_{A}(x)=n$, we have
  $n=\sum_{i=1}^{k} m_{i}$. By assumption, $m_i=g_i$  for all~$i$.  So it
  follows that the sum of the geometric multiplicities is~$n$, and hence,~$A$ is
  diagonalizable.
\end{proof}

\textbf{Jordan form.} What can we say when a linear transformation is not diagonalizable?  Can we
still choose a basis to make the matrix for the transformation simple in some
sense?  We give one answer here.  First, we need a couple definitions.  A
\emph{Jordan block of size~$k$ for~$\lambda\in F$} is the~$k\times k$
matrix with~$\lambda$s on the diagonal and~$1$s on the ``superdiagonal'':
\[
  J_{k}(\lambda)=
  \left(\begin{array}{ccccccc}
      \lambda&1&0&0&\cdots&0&0\\
      0&\lambda&1&0&\cdots&0&0\\
      0&0&\lambda&1&\cdots&0&0\\
      &&\vdots&&\ddots&\vdots&\\
      0&0&0&0&\cdots&\lambda&1\\
      0&0&0&0&\cdots&0&\lambda
  \end{array} \right)
\]
For example,
\[
  J_4(3)=
  \left(\begin{array}{cccc}
      3&1&0&0\\
      0&3&1&0\\
      0&0&3&1\\
      0&0&0&3\\
  \end{array} \right).
\]
Note the following example of the special case of a Jordan block of
size~$1$:
\[
  J_1(5)=[5].
\]

A matrix is in \emph{Jordan form} if it is in block diagonal form with
Jordan blocks for various~$\lambda$ along the diagonal:
\[
  \left(\begin{array}{cccc}
      J_{k_1}(\lambda_1)&0&\cdots&0\\
      0&J_{k_2}(\lambda_2)&\cdots&0\\
      0&0&\ddots&0\\
      0&0&\cdots&J_{k_m}(\lambda_m)
  \end{array} \right)
\]
For example, here is a matrix in Jordan form:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {
      $
  \left(\begin{array}{cccccccc}
      2&1&0&0&0&0&0&0\\
      0&2&0&0&0&0&0&0\\
      0&0&2&1&0&0&0&0\\
      0&0&0&2&0&0&0&0\\
      0&0&0&0&5&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
  \end{array} \right).
  $
  };
  \fill[fill=blue!60,opacity=0.2] (-2.16,1.73) rectangle (-1.20,0.85);
  \fill[fill=blue!60,opacity=0.2] (-1.11,0.9) rectangle (-0.15,0.03);
  \fill[fill=blue!60,opacity=0.2] (-0.0,0.03) rectangle (0.37,-0.4);
  \fill[fill=blue!60,opacity=0.2] (0.5,-0.41) rectangle (2.0,-1.65);
\end{tikzpicture} 
\end{center}
It has two~$2\times 2$ Jordan blocks for~$2$, a~$1\times 1$ Jordan block
for~$5$, and a~$3\times 3$ Jordan block for~$4$:
\[
  \left(\begin{array}{cccc}
      J_2(2)&0&0&0\\
      0&J_2(2)&0&0\\
      0&0&J_1(5)&0\\
      0&0&0&J_3(4)
  \end{array} \right).
\]
\medskip

\textbf{Theorem.} Let~$\dim V<\infty$.  Suppose~$f\colon V\to V$ is a linear transformation
over~$F$ and that the characteristic polynomial for~$f$ splits, i.e., the
field~$F$ contains all of the zeros of the characteristic polynomial.  Then
there exists an ordered basis for~$V$ such that the matrix
representing~$f$ with respect to that basis is in Jordan form.  The Jordan form
is unique up to a permutation of the Jordan blocks.
\medskip

So a matrix is diagonalizable if and only if its characteristic polynomial
splits and all of its Jordan blocks have size~$1$.  We also know that a matrix
such as
\[
  \left(\begin{array}{ccc}
      5&1&0\\
      0&5&0\\
      0&0&2
  \end{array} \right),
\]
which is already in Jordan form but not diagonal, is not diagonalizable.
\end{document}  
