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\noindent \hfill
\fbox{\fbox{\parbox{7.25in}{\scshape{Math 2240 \hfill Eigenhandout}}} \hfill }

\vspace{0.2in}

\noindent Let $V$ be a vector space over $\mathbb{R}$ (or $\mathbb{C}$) such that $\dim(V)=n<\infty$. Let
          $T:V \rightarrow V$ be a linear transformation (since we are mapping from $V$ 
          to itself, we could refer to $T$ as a linear operator).

\vspace{0.1in}

\noindent {\bf Definition:} Let $\vv \in V$ such that $\vv \not= {\bf 0}$ and
$T(\vv)=\lambda \vv$. Then $\vv$ is an {\it eigenvector} for $T$ with 
{\it eigenvalue} $\lambda$. Moreover, we say that $\lambda \in \mathbb{R}$ (or $\mathbb{C}$)
is an {\it eigenvalue} for $T$ if $T$ has an eigenvector with eigenvalue
$\lambda$.

\vspace{0.1in}

{\it Note:} While $0$ can be an eigenvalue, ${\bf 0}$ is not allowed to be an eigenvector. Otherwise, since $T({\bf 0}) = {\bf 0} = \lambda {\bf 0}$, we would have that every scalar is an eigenvalue of $T$ and ${\bf 0}$ would have every scalar as its eigenvalue!

\vspace{0.1in}

\noindent
{\bf Definition:} Let $f(t) = \det(T-tI)$. Then $f(t)$ is called the
{\it characteristic polynomial} of $T$.\footnote{This is the definition in Lay. Other texts define $g(t)=\det(tI-T)$ to be the characteristic polynomial. Notice that $f(t)=(-1)^ng(t)$ where $n=\dim(V)$. So for even sized matrices $f=g$ and for odd sized matrices $f=-g$.}

\vspace{0.1in}

{\it Note:} $\lambda$ is an eigenvalue of $T$ $\Leftrightarrow$ there exists
an non-zero vector $\vv$ such that $T(\vv)=\lambda\vv$ $\Leftrightarrow$
there exists a non-zero vector $\vv$ such that $(T-\lambda I)(\vv)={\bf 0}$
$\Leftrightarrow$ $\mathrm{Ker}(T-\lambda I) \not= \{{\bf 0}\}$
$\Leftrightarrow$ $T-\lambda I$ is not 1-1
$\Leftrightarrow$ $T-\lambda I$ is not invertible
$\Leftrightarrow$ $\det(T-\lambda I)\not= 0$.
We have just proved...

\vspace{0.1in}

\noindent 
{\bf Theorem:} $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a root
of the characteristic polynomial of $T$ (that is $f(\lambda) = \det(T-\lambda I)=0$).

\vspace{0.1in}

\noindent 
{\bf Facts:} Let $f(t)$ be the characteristic polynomial of $T$. Then $f(t)$ is 
a polynomial of degree $n$ whose leading coefficient is $(-1)^n$. In addition, $f(0) = \det(T)$ (the constant term is just the determinant of $T$). 
Also, the coefficient of $t^{n-1}$ in $f(t)$ is $(-1)^{n-1}\mathrm{tr}(T)$ (i.e., $\pm$ the trace of $T$).

\vspace{0.1in}
\noindent 
{\bf Definition:} Factor $T$'s characteristic polynomial (over $\mathbb{C}$): 
$f(t) = (-1)^n(t-\lambda_1)^{m_1}(t-\lambda_2)^{m_2}\cdots(t-\lambda_k)^{m_k}$ (where
$\lambda_i \not= \lambda_j$ for $i \not= j$ and $m_i > 0$). Then the roots 
of $f(t)$ (i.e. the eigenvalues of $T$) are $\lambda_1,\dots,\lambda_k$.
We say that the {\it algebraic multiplicity} of $\lambda_i$ is $m_i$ (the
number of factors $(t-\lambda_i)$ appearing in the characteristic polynomial).
Notice that the sum of the algebraic multiplicities is $n = \dim(V)$ (the degree of the
characteristic polynomial).\\ {\it Technical note:} More accurately, if we are working over $\mathbb{R}$, the non-real roots are not actually eigenvalues.

\vspace{0.1in}
\noindent
{\bf Definition:} Let $E_\lambda = \mathrm{Ker}(T-\lambda I) = \{ \vv \in V \;|\; (T-\lambda I)(\vv) = {\bf 0} \} = \{ \vv \in V \;|\; T(\vv) = \lambda\vv \} = 
\{ {\bf 0} \} \cup \{ \vv \in V \,|\, \vv \mathrm{\;is\;an\;eigenvector\;of\;}T\mathrm{\;with\;eigenvalue\;}\lambda \}$.
If $E_\lambda \not= \{ {\bf 0} \}$ (i.e., $\lambda$ is an eigenvalue), then we call $E_\lambda$ an {\it eigenspace} of 
$T$. Notice that $E_\lambda$ is a subspace of $V$ (since it is the kernel of a linear transformation).

\vspace{0.1in}

\noindent
{\bf Definition:} $\dim(E_\lambda) = \dim(\mathrm{Ker}(T-\lambda I)) = \mathrm{nullity}(T-\lambda I)$ is called the {\it geometric multiplicity} of $\lambda$.
This is the number of linearly independent eigenvectors with eigenvalue $\lambda$. Notice that if $\lambda$ is an eigenvalue then $E_\lambda$ cannot be 
the zero subspace. Thus geometric multiplicities of eigenvalues are always at least $1$.


\vspace{0.1in}
\noindent
{\bf Theorem:} Let $\lambda$ be an eigenvalue of $T$ with algebraic mult. $m$ and geometric mult. $g$. Then $1 \leq g \leq m$.

\vspace{0.1in}
\noindent
{\bf Theorem:} Eigenvectors with different eigenvalues are linearly independent.
Moreover, if $S_i$ is a linearly independent set of eigenvectors with eigenvalue $\lambda_i$
and $\lambda_i \not= \lambda_j$ for $i \not= j$, then $S_1 \cupdot S_2 \cupdot \cdots \cupdot S_k$
is a linearly independent set.

\vspace{0.1in}
\noindent
{\bf Definition:} $T$ is {\it diagonalizable} if there is a basis for $V$ consisting
of eigenvectors for $T$. Notice if $\beta$ is such a basis, then $[T]_\beta$ is
a diagonal matrix!

\vspace{0.1in}
\noindent 
{\bf Corollary:} $T$ is diagonalizable (over $\mathbb{R}$) if and only if the eigenvalues of $T$ all belong to $\mathbb{R}$ (i.e. the characteristic polynomial completely factors over $\mathbb{R}$) and the geometric and algebraic multiplicities of each eigenvalue match.



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