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7.1:  Eigenvalues and Eigenvectors

Defn:  $\lambda$ is an {\bf eigenvalue} of the linear transformation
\hfil \break
$T:V
\rightarrow V$ if there exists a \underbar{nonzero} vector ${\bf x}$ in $V$
such that $T({\bf x}) = \lambda {\bf x}$.  The vector ${\bf x}$ is said to be
an {\bf eigenvector} corresponding to the eigenvalue $\lambda$.


Example:
Let $T({\bf x}) = \left[\matrix{4 & 1 \cr 5 & 0}\right] {\bf x}$.

\vskip 10pt
\hrule

 Note $\left[\matrix{4 & 1 \cr 5 & 0}\right] \left[\matrix{ -1 \cr ~~5
}\right] = \left[\matrix{ ~~1 \cr -5 }\right] = -1\left[\matrix{ -1 \cr ~~5
}\right]$

Thus -1 is an eigenvalue of $A$ and $\left[\matrix{ -1 \cr ~~5 }\right]$ is a
corresponding eigenvector of $A$.

\vskip 10pt
\hrule
 Note $\left[\matrix{4 & 1 \cr 5 & 0}\right] \left[\matrix{
1 \cr 1 }\right] =
\left[\matrix{ 5 \cr 5 }\right] = 5\left[\matrix{ 1 \cr 1 }\right]$

Thus 5 is an eigenvalue of $A$ and $\left[\matrix{ 1 \cr 1 }\right]$ is a
corresponding eigenvector of $A$.

\vskip 10pt
\hrule
 Note $\left[\matrix{4 & 1 \cr 5 & 0}\right] \left[\matrix{
2 \cr 8 }\right] =
\left[\matrix{ 16 \cr 10 }\right] \not= k\left[\matrix{ 2 \cr 8 }\right]$ for
any $k$.

Thus $\left[\matrix{ 2 \cr 8 }\right]$ is \underbar{NOT} an
eigenvector of $A$.

\eject
MOTIVATION:

Note $\left[\matrix{ 2 \cr 8 }\right] = \left[\matrix{ -1 \cr ~~5 }\right] +
3\left[\matrix{ 1 \cr 1 }\right]$

Thus $A\left[\matrix{ 2 \cr 8 }\right] = A(\left[\matrix{ -1 \cr ~~5 }\right]
+ 3\left[\matrix{ 1 \cr 1 }\right]) = A\left[\matrix{ -1 \cr ~~5 }\right] +
3A\left[\matrix{ 1 \cr 1 }\right]$

\vskip 5pt
\rightline{$= -1\left[\matrix{ -1 \cr ~~5 }\right] + 3
\cdot 5\left[\matrix{ 1 \cr 1 }\right] = \left[\matrix{ 16 \cr 10 }\right] $}

\vskip 30pt

Finding eigenvalues:

Suppose $A{\bf x} = \lambda {\bf x}$ ~~(Note $A$ is a SQUARE matrix).

Then $A{\bf x} = \lambda I{\bf x}$ where $I$ is the identity matrix.

Thus $\lambda I{\bf x} - A{\bf x} = (\lambda I - A){\bf x} = {\bf 0}$

Thus if $A{\bf x} = \lambda {\bf x}$ for a nonzero {\bf x}, then $(\lambda I- A){\bf x}
= {\bf 0}$ has a nonzero solution. (mod 11/14)

Thus $det(\lambda I - A){\bf x} = 0$.

Note that the eigenvectors corresponding to $\lambda$ are the nonzero
solutions of $(\lambda I - A){\bf x} = {\bf 0}$.

\eject
Thus to find the eigenvalues of $A$ and their corresponding eigenvectors:

Step 1: Find eigenvalues:  Solve the equation 
\vskip 4pt
\centerline{$det(\lambda I - A) = 0$
for $\lambda$.}

Step 2:  For each eigenvalue $\lambda_0$, find its corresponding eigenvectors
by
solving the homogeneous system of equations 
\vskip 4pt

\centerline{$(\lambda_0 I - A){\bf x} = 0$
for {\bf x}.  (mod 11/15)
}

\vskip 5pt
\hrule
\vskip -3pt

Defn:  $det(\lambda I - A) = 0$ is the {\bf characteristic equation} of
$A$.


Thm 3:  The eigenvalues of an upper triangular or lower triangular matrix
(including diagonal matrices) are identical to its diagonal entries.

\vskip 5pt
\hrule
\vskip -3pt

Defn:  The {\bf eigenspace} corresponding to an eigenvalue $\lambda_0$
of a matrix $A$ is
the set of all solutions of $(\lambda_0 I - A){\bf x} = {\bf 0}$.
(mod 11/15)
\vfill
Note:  An eigenspace is a vector space

\hskip 0.4in The vector {\bf 0} is always in the eigenspace.

\hskip 0.4in The vector {\bf 0} is never an eigenvector.

\hskip 0.4in The number 0 can be an eigenvalue.


Thm 7.1.4:  A square matrix is invertible if and only if $\lambda = 0$
is not an eigenvalue of $A$.

Thm 7.1.3:  If $A{\bf x} = \lambda{\bf x}$, then $A^k{\bf x} =
\lambda^k{\bf x}$. ~~ That is, if~\break  $\lambda$ is an eigenvalue
of
$A$
with corresponding eigenvector ${\bf x}$, then $\lambda^k$ is an
eigenvalue of $A^k$
with corresponding eigenvector ${\bf x}$ where $k$ is any
 integer.
\end