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\begin{document}

\title{Math 413/513 Chapter 5 (from Friedberg, Insel, \& Spence)}
\author{David Glickenstein}
\maketitle

\section{Eigenvalues and Eigenvectors}

\begin{definition}
If $V$ is a finite dimensional vector space, a linear operator $T\in
\mathcal{L}\left(  V\right)  $ is \emph{diagonalizable} if there is an ordered
basis $\beta$ for $V$ such that $\left[  T\right]  _{\beta}$ is a diagonal
matrix. A square matrix $A$ is diagonalizable if $L_{A}$ is
\emph{diagonalizable}.
\end{definition}

This maybe requires some basic examples. Notice that
\[
T\left(  x,y\right)  =\left(  x+y,4x+y\right)
\]
does not produce a diagonal matrix in the standard basis
\[
\left[  T\right]  _{E_{2}}=\left(
\begin{array}
[c]{cc}%
1 & 1\\
4 & 1
\end{array}
\right)
\]
but if we consider the basis
\[
\beta=\left\{  \left(  1,2\right)  ,\left(  1,-2\right)  \right\}
\]
we see that
\begin{align*}
T\left(  1,2\right)   &  =\left(  3,6\right)  =3\left(  1,2\right) \\
T\left(  1,-2\right)   &  =\left(  -1,2\right)  =-\left(  1,-2\right)
\end{align*}
so
\[
\left[  T\right]  _{\beta}=\left(
\begin{array}
[c]{cc}%
3 & 0\\
0 & -1
\end{array}
\right)  .
\]
Hence $T$ is diagonalizable and so is $\left(
\begin{array}
[c]{cc}%
1 & 1\\
4 & 1
\end{array}
\right)  .$

Notice that the fact that $\left[  T\right]  _{\beta}$ is diagonal is
equivalent to the fact that $T\left(  v_{1}\right)  =\lambda_{1}v_{1}$ for an
ordered basis $\left\{  v_{1},\ldots,v_{n}\right\}  ,$ where $\lambda
_{1},\ldots,\lambda_{n}$ are scalars. The choice of $v_{i}$ is important here,
since if we choose other vectors, $T$ does not have this simple form (in the
above example, $T\left(  1,0\right)  \neq\lambda\left(  1,0\right)  $ for any
choice of scalar $\lambda$). This leads to the notion of eigenvalue and eigenvector.

\begin{definition}
Let $T$ be a linear operator on a vector space $V$ (not necessarily finite
dimensional). A nonzero vector $v\in V$ is called an\emph{ eigenvector} of $T$
if there exists a scalar $\lambda$ such that $T\left(  v\right)  =\lambda v.$
The scalar $\lambda$ is called the \emph{eigenvalue }corresponding to the
eigenvector $v.$ Let $A\in F^{n\times n}.$ A nonzero vector $v\in F^{n}is$
called an eigenvector of $A$ if $v$ is an eigenvector of $L_{A},$ i.e.,
$Av=\lambda v$ for some $\lambda.$ The scalar $\lambda$ is called the
eigenvalue corresponding to the eigenvector $v.$
\end{definition}

\begin{remark}
Some older terms for eigenvalue/eigenvector are characteristic
value/characteristic vector and proper value/proper vector.
\end{remark}

\begin{theorem}
A vector $v$ is an eigenvector for a linear operator $T\in\mathcal{L}\left(
V\right)  $ corresponding to eigenvalue $\lambda$ if and only if $v\neq0$ and
$v\in N\left(  T-\lambda I_{V}\right)  .$
\end{theorem}

\begin{theorem}
A linear operator $T$ on a finite-dimensional vector space $V$ is
diagonalizable if and only if there exists an ordered basis $\beta$ for $V$
consisting of eigenvectors of $T.$ Furthermore, if $T$ is diagonalizable and
$\beta$ is an ordered basis of eigenvectors of $T$ and $D=\left[  T\right]
_{\beta},$ then $D$ is a diagonal matrix and the diagonal entries are the
corresponding eigenvalues.
\end{theorem}

The idea of the proof is essentially the discussion above and is left as an exercise.

\begin{example}
Not all matrices have eigenvectors. We can determine rotation by angle
$\theta$ by left multiplication by the matrix
\[
R_{\theta}=\left(
\begin{array}
[c]{cc}%
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{array}
\right)  .
\]
If $\theta$ is not an integer multiple of $\pi,$ then $R_{\theta}$ does not
take any nonzero vector to a multiple of itself, and so these matrices have no eigenvectors.
\end{example}

\begin{example}
Let $C^{\infty}$ be the vector space consisting of function
$\mathbb{R\rightarrow\mathbb{R}}$ with derivatives of all orders (check this
is a vector space subspace of the vector space of functions
$\mathbb{R\rightarrow\mathbb{R}}$). Consider the map $T:C^{\infty}\left(
\mathbb{R}\right)  \rightarrow C^{\infty}\left(  \mathbb{R}\right)  $ given by
$T\left(  f\right)  =f^{\prime}.$ For a function to be an eigenvector, we need
that $f^{\prime}=\lambda f$ for some value of $\lambda.$ There are functions
that satisfy this property, namely $f_{\lambda}\left(  x\right)  =e^{\lambda
x}$ is a eigenvector with eigenvalue $\lambda.$ Note that the constant
function is an eigenvector with eigenvalue $0.$ 
\end{example}

Notice that if $v$ is an eigenvector for a linear operator $T$ with
corresponding eigenvalue $\lambda,$ then $T\left(  v\right)  =\lambda v$ can
be written $\left(  T-\lambda I_{V}\right)  v=0$ where $I_{V}$ is the identity
transformation on $V.$ Thus there is a nonzero element of the nullspace of the
linear operator $T-\lambda I.$ For matrices, this allows us to use
determinants to find eigenvalues.

\begin{theorem}
Let $A\in F^{n\times n}.$ The scalar $\lambda$ is an eigenvalue of $A$ if and
only if $\det\left(  A-\lambda I_{n}\right)  =0.$
\end{theorem}

\begin{proof}
If $\lambda$ is an eigenvalue, its corresponding eigenvector satisfies
$\left(  A-\lambda I_{n}\right)  v=0,$ so there $A-\lambda I_{n}$ is not
invertible, so $\det\left(  A-\lambda I_{n}\right)  =0.$ Similarly, if
$\det\left(  A-\lambda I_{n}\right)  =0,$ then $A-\lambda I_{n}$ is not
invertible, so there is a nonzero element $v$ in the nullspace of $A-\lambda
I_{n},$ and then $Av=\lambda v.$
\end{proof}

One can replace $\lambda$ in the previous theorem to get a polynomial
associated with a matrix $A.$

\begin{definition}
Let $A\in F^{n\times n}.$ The polynomial $f\left(  t\right)  =\det\left(
A-tI_{n}\right)  $ is called the \emph{characteristic polynomial} of $A.$
\end{definition}

Note that the theorem above says that any eigenvalue is a zero of the
characteristic polynomial and any zero of the characteristic polynomial is an
eigenvalue. We can use this to compute eigenvalues, then eigenvectors, and
sometimes even diagonalize the matrix.

Suppose we have a basis of eigenvectors for a square matrix $A$, then notice
that if $Q$ has the elements of the basis as its columns, then $AQ=QD$ where
$D$ is the diagonal matrix with the eigenvalues on the diagonal, hence
$Q^{-1}AQ=D.$ 

\begin{example}
Consider the matrix
\[
A=\left(
\begin{array}
[c]{ccc}%
1 & 4 & 5\\
0 & 2 & 6\\
0 & 0 & 3
\end{array}
\right)  .
\]
The characteristic polynomial is
\[
\left(  1-\lambda\right)  \left(  2-\lambda\right)  \left(  3-\lambda\right)
.
\]
Each must correspond to at least one eigenvector, so we need to calculate%
\[
A-I=\left(
\begin{array}
[c]{ccc}%
0 & 4 & 5\\
0 & 1 & 6\\
0 & 0 & 2
\end{array}
\right)
\]
and we see that $N\left(  A-I\right)  =\operatorname{span}\left\{  \left(
1,0,0\right)  \right\}  .$ We calculate
\[
A-2I=\left(
\begin{array}
[c]{ccc}%
-1 & 4 & 5\\
0 & 0 & 6\\
0 & 0 & 1
\end{array}
\right)
\]
and so $N\left(  A-2I\right)  =\operatorname{span}\left\{  \left(
4,1,0\right)  \right\}  .$ We calculate
\[
A-3I=\left(
\begin{array}
[c]{ccc}%
-2 & 4 & 5\\
0 & -1 & 6\\
0 & 0 & 0
\end{array}
\right)
\]
and so
\[
N\left(  A-3I\right)  =\operatorname{span}\left\{  \left(  29,12,2\right)
\right\}  .
\]
Let's double check that these are, in fact, eigenvectors%
\begin{align*}
\left(
\begin{array}
[c]{ccc}%
1 & 4 & 5\\
0 & 2 & 6\\
0 & 0 & 3
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
1\\
0\\
0
\end{array}
\right)    & =\left(
\begin{array}
[c]{c}%
1\\
0\\
0
\end{array}
\right)  \\
\left(
\begin{array}
[c]{ccc}%
1 & 4 & 5\\
0 & 2 & 6\\
0 & 0 & 3
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
4\\
1\\
0
\end{array}
\right)    & =\left(
\begin{array}
[c]{c}%
8\\
2\\
0
\end{array}
\right)  =2\left(
\begin{array}
[c]{c}%
4\\
1\\
0
\end{array}
\right)  \\
\left(
\begin{array}
[c]{ccc}%
1 & 4 & 5\\
0 & 2 & 6\\
0 & 0 & 3
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
29\\
12\\
2
\end{array}
\right)    & =\left(
\begin{array}
[c]{c}%
87\\
36\\
6
\end{array}
\right)  =3\left(
\begin{array}
[c]{c}%
29\\
12\\
2
\end{array}
\right)  .
\end{align*}
One can check that
\[
Q=\left(
\begin{array}
[c]{ccc}%
1 & 4 & 29\\
0 & 1 & 12\\
0 & 0 & 2
\end{array}
\right)
\]
has inverse
\[
Q^{-1}=\left(
\begin{array}
[c]{ccc}%
1 & -4 & \frac{19}{2}\\
0 & 1 & -6\\
0 & 0 & \frac{1}{2}%
\end{array}
\right)
\]
and
\[
\left(
\begin{array}
[c]{ccc}%
1 & -4 & \frac{19}{2}\\
0 & 1 & -6\\
0 & 0 & \frac{1}{2}%
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
1 & 4 & 5\\
0 & 2 & 6\\
0 & 0 & 3
\end{array}
\right)  \left(
\begin{array}
[c]{ccc}%
1 & 4 & 29\\
0 & 1 & 12\\
0 & 0 & 2
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{array}
\right)
\]

\end{example}

We can see how similar matrices affect the characteristic polynomial.

\begin{proposition}
If $A$ is similar to $B,$ that is, there exists an invertible matrix $Q$ such
that $A=Q^{-1}BQ,$ then the characteristic polynomials of $A$ and $B$ are the same.
\end{proposition}

\begin{proof}
Exercise.
\end{proof}

Recall that if $\beta,\beta^{\prime}$ are ordered bases for a
finite-dimensional vector space $V$, then there exists and invertible matrix
$Q$ such that for any $T\in\mathcal{L}\left(  V\right)  $ we have $\left[
T\right]  _{\beta}=Q^{-1}\left[  T\right]  _{\beta^{\prime}}Q.$ Thus, we can
define the characteristic polynomial of a linear transformation.

\begin{definition}
Let $V$ be a finite-dimensional vector space and let $T\in\mathcal{L}\left(
V\right)  .$ We define the characteristic polynomial of $T$ to be the
characteristic polynomial of $\left[  T\right]  _{\beta}$ for any basis
$\beta$ of $V,$ i.e., the characteristic polynomial is $f\left(  t\right)
=\det\left(  \left[  T\right]  _{\beta}-\lambda I_{n}\right)  $ where $n=\dim
V.$
\end{definition}

\begin{remark}
Note that $\left[  T\right]  _{\beta}-\lambda I_{n}=\left[  T-\lambda
I_{V}\right]  _{\beta}$ so we could have defined it thus.
\end{remark}

It should be fairly clear from the expansion in cofactors formula that the
characteristic polynomial is, in fact, a polynomial. We can be a bit more precise.

\begin{theorem}
Let $A\in F^{n\times n}.$ The characteristic polynomial of $A$ is a polynomial
of degree $n$ with leading coefficient $\left(  -1\right)  ^{n}.$ Also, $A$
has at most $n$ distinct eigenvalues.
\end{theorem}

\begin{proof}
Exercise.
\end{proof}

\section{Problems}

\begin{itemize}
\item Be sure to read more examples in the book on diagonalizing matrices by
finding zeroes of the characteristic polynomial and then finding eigenvectors.

\item FIS Section 5.1 exercises 2, 3, 7, 8ab, 9 12, 14, 15, 16, 19-22
\end{itemize}

\section{Diagonalizability}

\begin{theorem}
\label{thm:distinct eigenvalues}Let $T$ be a linear operator on a vector space
$V$ and let $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ be distinct
eigenvalues of $T.$ If $v_{1},\ldots,v_{k}$ are eigenvectors of $T$ such that
$T\left(  v_{i}\right)  =\lambda_{i}v_{i}$ for $i=1,\ldots,k,$ then $\left\{
v_{1},\ldots,v_{k}\right\}  $ is linearly independent.
\end{theorem}

\begin{proof}
Consider a linear combination
\[
\sum_{i=1}^{k}a_{i}v_{i}=\vec{0}.
\]
It follows by applying $T-\lambda_{k}I_{V}$ that
\[
\sum_{i=1}^{k-1}a_{i}\left(  \lambda_{i}-\lambda_{k}\right)  v_{i}=\vec{0}.
\]
We can now use this to induct on $k.$ If $k=1,$ then since $v_{1}\neq\vec{0}$
(since eigenvectors are nonzero), $\left\{  v_{1}\right\}  $ is linearly
independent. Now suppose any set of eigenvectors of distinct eigenvalues that
has fewer than $k$ elements is linearly independent. We see that if
\[
\sum_{i=1}^{k}a_{i}v_{i}=\vec{0}%
\]
then we can apply $T-\lambda_{k}I_{V}$ to both sides to find that
\[
\sum_{i=1}^{k-1}a_{i}\left(  \lambda_{i}-\lambda_{k}\right)  v_{i}=\vec{0}.
\]
Since $\left\{  v_{1},\ldots,v_{k-1}\right\}  $ are linearly independent and
$\lambda_{i}\neq\lambda_{k},$ we must have that $a_{1}=\cdots=a_{k-1}=0.$ It
then follows that $a_{k}=0$ as well. 
\end{proof}

\begin{corollary}
Let $V$ be an $n$-dimensional vector space and $T\in\mathcal{L}\left(
V\right)  .$ If $T$ has $n$ distinct eigenvalues, then $T$ is diagonalizable.
\end{corollary}

\begin{proof}
$T$ has at least one eigenvector for each eigenvalue, yielding a set $\left\{
v_{1},\ldots,v_{n}\right\}  $. By the theorem, this set is linearly
independent, and thus a basis. By a previous theorem, if $V$ has a basis of
eigenvectors of $T,$ then $T$ is diagonalizable.
\end{proof}

Note that the converse of the theorem is false; we can have a linear operator
that is diagonalizable that has repeated eigenvalues (for instance, the
identity map).

\begin{definition}
A polynomial $f\left(  t\right)  $ in $P\left(  F\right)  $ splits over $F$ if
there are scalars $c,a_{1},\ldots,a_{n}$ (not necessarily distinct) in $F$
such that
\[
f\left(  t\right)  =c\left(  t-a_{1}\right)  \left(  t-a_{2}\right)
\cdots\left(  t-a_{n}\right)  .
\]

\end{definition}

So basically a polynomial splits if it can be factored with all linear
factors. Note that the field is important. For instance, $t^{2}+1$ does not
split over $\mathbb{R}$ or $\mathbb{Q}$ but it does split over $\mathbb{C}$
since $t^{2}+1=\left(  t+i\right)  \left(  t-i\right)  .$ Also note that we
did not preclude $\left(  t-1\right)  ^{2},$ which also splits.

\begin{theorem}
The characteristic polynomial of any diagonalizable linear operator splits.
\end{theorem}

\begin{proof}
It is clear that if $D$ is a diagonal matrix, then its characteristic
polynomial splits. If $T$ is diagonalizable, then there is a basis $\beta$
such that $\left[  T\right]  _{\beta}$ is diagonal, so the theorem follows.
\end{proof}

If we think a bit harder about the proof, we see that it may be possible to
have an eigenvalue repeated when written in diagonal form, and this leads to a
repeated zero in the characteristic polynomial. 

\begin{definition}
Let $\lambda$ be an eigenvalue of a linear operator or matrix with
characteristic polynomial $f\left(  t\right)  .$ The (algebraic)\ multiplicity
of $\lambda$ is the largest positive integer $k$ such that $\left(
t-\lambda\right)  ^{k}$ is a factor of $f\left(  t\right)  .$
\end{definition}

While we have shown that if a transformation is diagonalizable and a diagonal
representation has a repeated eigenvalue, then their must be a repeated zero
in the characteristic polynomial. It is natural to ask if a transformation
whose characteristic polynomial splits with a repeated eigenvalue must be
diagonalizable, but this is false.

\begin{example}
Consider the matrix
\[
A=\left(
\begin{array}
[c]{cc}%
0 & 1\\
0 & 0
\end{array}
\right)  .
\]
Its characteristic polynomial is $\lambda^{2},$ so it splits. We see that
$N\left(  A\right)  =\operatorname{span}\left\{  \left(  1,0\right)  \right\}
,$ which is one-dimensional. It is therefore impossible to have a basis of
eigenvectors and so $A$ is not diagonalizable.
\end{example}

The problem in the last example is that we are unable to find a basis of
eigenvectors since there is only one eigenvalue and the nullspace of
$A-\lambda I_{V}$ (where $\lambda=0$) is only one-dimensional. This leads to
the following definition.

\begin{definition}
Let $T$ be a linear operator on a vector space $V$ and let $\lambda$ be an
eigenvalue of $T.$ Define $E_{\lambda}=\left\{  x\in V:T\left(  x\right)
=\lambda x\right\}  =N\left(  T-\lambda I_{V}\right)  .$ The set $E_{\lambda}$
is called the \emph{eigenspace} of $T$ corresponding to the eigenvalue
$\lambda.$ Analogously, the eigenspace of a matrix $A$ is the eigenspace of
$L_{A}.$
\end{definition}

\begin{theorem}
Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let
$\lambda$ be an eigenvalue of $T$ with multiplicity $m.$ Then $1\leq\dim
E_{\lambda}\leq m.$
\end{theorem}

\begin{proof}
Choose an ordered basis $\left\{  v_{1},\ldots,v_{p}\right\}  $ for
$E_{\lambda},$ then extend to a basis $\beta$ for $V.$ Then $\left[  T\right]
_{\beta}$ will have the block form%
\[
\left(
\begin{array}
[c]{cc}%
\lambda I_{p} & A\\
0 & B
\end{array}
\right)  .
\]
It is clear that the characteristic polynomial will look like $f\left(
t\right)  =\left(  \lambda-t\right)  ^{p}\det\left(  B-tI\right)  .$ It
follows that the multiplicity $m$ is at least as big as $p=\dim E_{\lambda}.$
Since $\lambda$ is an eigenvalue, it has at least one eigenvector so $\dim
E_{\lambda}\geq1.$
\end{proof}

\begin{theorem}
\label{thm:basis}Let $T$ be a linear operator on a vector space $V$ and let
$\lambda_{1},\ldots,\lambda_{k}$ be distinct eigenvalues of $T.$ For each
$i=1,2,\ldots,k$ let $S_{i}$ be a finite linearly independent subset of
$E_{\lambda_{i}}.$ Then $S=S_{1}\cup S_{2}\cup\cdots\cup S_{k}$ is a linearly
independent subset of $V.$ [Note: the union should really be more of a
concatenation since it needs to preserve the ordering of the $S_{i}$.]
\end{theorem}

\begin{proof}
We need some notation to keep track of the vectors in $S_{i}$ and $S.$ Suppose
$S_{i}=\left\{  v_{i1},\ldots,v_{in_{i}}\right\}  $ for each $i$ and that
$S=\left\{  v_{11},\ldots,v_{1n_{1}},v_{21},\ldots,v_{kn_{k}}\right\}  .$
Consider scalars $a_{ij}$ such that
\[
\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}a_{ij}v_{ij}=0.
\]
For each $i,$ let $w_{i}=\sum_{j=1}^{n_{i}}a_{ij}v_{ij}.$ Note that $w_{i}\in
S_{i}$ and $\sum_{i=1}^{k}w_{i}=0.$ If we could show that this implies
$w_{i}=0$ for all $i=1,\ldots,k,$ then it will follow that $a_{ij}=0$ for all
$i,j$ (since $\left\{  v_{i1},\ldots,v_{in_{i}}\right\}  $ is a basis for each
$i$).This is the content of the following lemma.
\end{proof}

\begin{lemma}
Let $T$ be a linear operator and let $\lambda_{1},\ldots,\lambda_{p}$ be
distinct eigenvalues of $T.$ For each $i=1,\ldots,p,$ let $v_{i}\in
E_{\lambda_{i}}.$ If
\[
v_{1}+\cdots+v_{p}=\vec{0}%
\]
then $v_{i}=0$ for all $i.$
\end{lemma}

\begin{proof}
By Theorem \ref{thm:distinct eigenvalues} , $N=\left\{  v_{i}:v_{i}\neq\vec
{0}\right\}  $ is a linearly independent set. Thus, if $N$ is nonempty, then
$v_{1}+\cdots+v_{p}\neq\vec{0}.$
\end{proof}

\begin{theorem}
Let $T$ be a linear operator on a finite-dimensional vector space $V$ and let
$\lambda_{1},\ldots,\lambda_{k}$ be the distinct eigenvalues of $T.$ then

\begin{enumerate}
\item $T$ is diagonalizable if and only if the characteristic polynomial of
$T$ splits and the multiplicity of $\lambda_{i}$ is equal to $\dim
E_{\lambda_{i}}$ for all $i.$

\item If $T$ is diagonalizable and $\beta_{i}$ is an ordered basis for
$E_{\lambda_{i}}$ for each $i,$ then $\beta=\beta_{1}\cup\beta_{2}\cup
\cdots\cup\beta_{k}$ is an ordered basis for $V$ consisting of eigenvectors of
$T.$
\end{enumerate}
\end{theorem}

\begin{proof}
If $T$ is diagonalizable, then there is a basis $\beta$ such that $\left[
T\right]  _{\beta}$ takes the block form%
\[
\left[  T\right]  _{\beta}=\left(
\begin{array}
[c]{cccc}%
\lambda_{1}I_{n_{1}} & 0 & \cdots & 0\\
0 & \lambda_{2}I_{n_{2}} & \ddots & \vdots\\
\vdots & \ddots & \ddots & 0\\
0 & \cdots & 0 & \lambda_{k}I_{n_{k}}%
\end{array}
\right)  .
\]
(Technically, there is a basis such that the matrix is diagonal, but then we
can reorder that basis so that equal eigenvalues are adjacent.) It follows
that the characteristic polynomial of $T$ is $\left(  \lambda_{1}-t\right)
^{n_{1}}\cdots\left(  \lambda_{k}-t\right)  ^{n_{k}}$ where $n_{1}%
,\ldots,n_{k}$ are the dimensions of $E_{\lambda_{1}},\ldots,E_{\lambda_{k}}.$

Now suppose the characteristic polynomial of $T$ splits is equal to $\left(
\lambda_{1}-t\right)  ^{n_{1}}\cdots\left(  \lambda_{k}-t\right)  ^{n_{k}}$
where $n_{1},\ldots,n_{k}$ are the dimensions of $E_{\lambda_{1}}%
,\ldots,E_{\lambda_{k}}.$ Then by Theorem \ref{thm:basis} we can take bases
$\beta_{i}$ of $E_{\lambda_{i}}$ and the set $\beta=\beta_{1}\cup\cdots
\cup\beta_{k}$ is linearly independent. Since it has size equal to
$n_{1}+\cdots+n_{k}=\dim V$ (since this is the degree of the characteristic
polynomial), $\beta$ must be a basis and in that basis, $\left[  T\right]
_{\beta}$ has the form above and is thus diagonalizable. Note that this proves
the second statement as well.
\end{proof}

\section{Problems}

\begin{itemize}
\item Be sure to read the examples in the book on diagonalizing.

\item FIS Section 5.2 exercises 2, 3, 6, 7 (diagonalize first: see Example 7),
8, 12, 13, 18, 19. 
\end{itemize}


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