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Eigenvalues.xml
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Eigenvalues.xml
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<chapter><title>Eigenvalues and eigenvectors</title>
<section><title>Eigenvalues and eigenvectors: preliminaries</title>
<p>
Eigenvalues and eigenvectors are defined for a square matrix <m>A.</m>
</p>
<definition><title>The eigenvalue of a matrix</title>
<statement>
<p>
A number <m>\lambda</m> is an <term>eigenvalue</term> of a square matrix <m>A</m>
if
<me>A\vec x=\lambda\vec x</me>
for some
<m>\vec x\not=\vec0</m>.
</p>
</statement>
</definition>
<definition><title>The eigenvector of a matrix</title>
<statement>
<p>
If
<me>
A\vec x=\lambda\vec x
</me>
for some <m>\vec x\not=\vec0</m>,
then <m>\vec x</m> is called an <term>eigenvector</term> of <m>A</m>
corresponding to the eigenvalue <m>\lambda</m>.
</p>
</statement>
</definition>
<p>
Notice that if <m>\vec x=\vec 0</m>, then <m>A\vec x=\lambda\vec x</m>
is simply the equation <m>\vec0=\vec0</m> for any value of <m>\lambda</m>.
This is not too interesting, and so we always have the restriction
<m>\vec x\not=\vec0</m>.
</p>
<example xml:id="EigenvalueExample1"><title>Eigenvalues of
<m>
A=
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix}
</m>
</title>
<p>
<me>
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix}
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
=
\begin{bmatrix}
2\\2\\2
\end{bmatrix}=2
\begin{bmatrix}
1\\1\\1
\end{bmatrix}
</me>
which makes
<m>\vec x=\begin{bmatrix}1\\1\\1\end{bmatrix}</m>
an eigenvector with eigenvalue
<m>\lambda=2</m>.
</p>
<p>
<me>
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix}
\begin{bmatrix}
1\\-1\\1
\end{bmatrix}
= \begin{bmatrix}
4\\-4\\4
\end{bmatrix}=4
\begin{bmatrix}
1\\-1\\1
\end{bmatrix}
</me>
which makes makes
<m>\vec
x=\begin{bmatrix}
1\\-1\\1
\end{bmatrix}
</m>
an eigenvector with eigenvalue <m>\lambda=4</m>.
</p>
<p>
<me>
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix}
\begin{bmatrix}
-1\\-1\\1
\end{bmatrix}=
\begin{bmatrix}
-6\\-6\\6
\end{bmatrix}=6
\begin{bmatrix}
-1\\-1\\1
\end{bmatrix}
</me>
which makes makes
<m>\vec x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}</m>
an eigenvector with eigenvalue <m>\lambda=6</m>.
</p>
<p>
We have now computed eigenvalues of <m>A</m>:
<m>\lambda=2</m>, <m>\lambda=4</m> and
<m>\lambda=6</m>.
With a little more theory, we will see that there are no others.
</p>
</example>
<example xml:id="EigenvalueExample2"><title>Eigenvalues of
<m>\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}</m>
</title>
<p>
Let
<m>A=\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}</m>
Then we have
<me>
\begin{bmatrix} 2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5\end{bmatrix}
\begin{bmatrix}1\\1\\0\end{bmatrix}= \begin{bmatrix}3\\3\\0\end{bmatrix}
=3 \begin{bmatrix}1\\1\\0\end{bmatrix}
</me>
which makes <m>\vec x=\begin{bmatrix}1\\1\\0\end{bmatrix}</m>
an eigenvector with eigenvalue <m>\lambda=3</m>.
Also
<me>
\begin{bmatrix} 2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5\end{bmatrix}
\begin{bmatrix}4\\0\\1\end{bmatrix}=
\begin{bmatrix}12\\0\\3\end{bmatrix}=3 \begin{bmatrix}4\\0\\1\end{bmatrix}
</me>
which makes
<m>\vec x=\begin{bmatrix}4\\0\\1\end{bmatrix}</m>
an eigenvector with eigenvalue <m>\lambda=3</m>
Finally
<me>
\begin{bmatrix} 2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5\end{bmatrix}
\begin{bmatrix}-1\\0\\2\end{bmatrix}=
\begin{bmatrix}6\\0\\-12\end{bmatrix}=-6 \begin{bmatrix}-1\\0\\2\end{bmatrix}
</me>
which makes
<m>\vec x=\begin{bmatrix}-1\\0\\2\end{bmatrix}</m>
an eigenvector with eigenvalue <m>\lambda=-6</m>
Hence the demonstrated eigenvalues are <m>\lambda=3</m> and <m>\lambda=-6</m>.
We will soon see that there are no more.
</p>
</example>
<example><title>Eigenvalues of
<m>\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}</m>
</title>
<p>
Let <m>A=\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}</m> and consider the equation
<m>A\vec x=\lambda\vec x</m>.
Setting <m>\vec x=\begin{bmatrix} x_1\\x_2\end{bmatrix}</m>,
<me>
\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}\begin{bmatrix} x_1\\x_2\end{bmatrix}
=\lambda\begin{bmatrix} x_1\\x_2\end{bmatrix}
</me>
This is equivalent to the system of equations
<me>
\begin{array}{rl}
\lambda x_1 \amp = -x_2\\
\lambda x_2 \amp = x_1.
\end{array}
</me>
This implies that
<m>(\lambda^2+1)x_1 = \lambda^2 x_1 + x_1 = -\lambda x_2 +\lambda x_2=0 </m>
and
<m>(\lambda^2+1)x_2 = \lambda^2 x_2 + x_2 = \lambda x_1 -\lambda x_21=0.</m>
Since <m>\vec x \not=\vec 0</m>, either <m>x_1\not=0</m> or <m>x_2\not=0</m>, and
consequently <m>\lambda^2=-1</m>. Since no real number satisfies this equation,
we conclude that there are no eigenvalues or eigenvectors for <m>A</m>.
An aside: if we consider complex numbers, then <m>i</m> and <m>-i</m> are both
eigenvalues of <m>A</m> with
<m>\begin{bmatrix} 1\\-i\end{bmatrix}</m>
and <m>\begin{bmatrix} 1\\i\end{bmatrix}</m> as corresponding eigenvectors.
</p>
</example>
<example xml:id="EigenvalueExample4"><title>Eigenvalues of
<m>\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}</m>
</title>
<p>
Consider the matrix
<me>
A=
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}
</me>
Then it's easy to verify that
<me>
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}
\begin{bmatrix} 1\\0\\0\\0\end{bmatrix}
=
\begin{bmatrix} 1\\0\\0\\0\end{bmatrix}
=
1\begin{bmatrix} 1\\0\\0\\0\end{bmatrix}
</me>
<me>
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}
\begin{bmatrix} 1\\1\\0\\0\end{bmatrix}
=
\begin{bmatrix} 2\\2\\0\\0\end{bmatrix}
=2\begin{bmatrix} 1\\1\\0\\0\end{bmatrix}
</me>
<me>
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}
\begin{bmatrix} 3\\4\\2\\0\end{bmatrix}
=
\begin{bmatrix} 9\\12\\6\\0\end{bmatrix}
=
3\begin{bmatrix} 3\\4\\2\\0\end{bmatrix}
</me>
<me>
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}
\begin{bmatrix} 8\\12\\9\\3\end{bmatrix}
=
\begin{bmatrix} 32\\48\\36\\12\end{bmatrix}
=4\begin{bmatrix} 8\\12\\9\\3\end{bmatrix}
</me>
and so we have <m>1</m>, <m>2</m>, <m>3</m> and <m>4</m> as eigenvalues.
Notice that in
this case the eigenvalues are just the diagonal elements and that
the matrix is upper triangular. We shall see
in <xref ref="EigenvaluesTriangularMatrix" />
that for every
upper (or lower) triangular matrix, the eigenvalues are the
diagonal entries.
</p>
</example>
<definition xml:id="EigenspaceDef"><title>Eigenspaces</title>
<statement>
<p>
Suppose that <m>A</m> is a square matrix of order <m>n</m>. Then for
any real number <m>\lambda</m>, we define the <term>eigenspace</term>
<m>E_\lambda</m> by
<me>E_\lambda=\{\vec x\in \R^n\mid A\vec x=\lambda \vec x\}.</me>
</p>
</statement>
</definition>
<p>
Clearly <m>\vec0</m> is in <m>E_\lambda</m> for any value of <m>\lambda</m>,
and <m>\lambda</m> is an eigenvalue if and only if there is some
<m>\vec x\not=\vec0</m> in <m>E_\lambda</m>.
</p>
<proposition>
<title>An eigenspace is a subspace</title>
<statement>
<p>
Let <m>A</m> be an <m>n\times n</m> matrix and let <m>\lambda</m> be a real number.
Then the set of vectors
<m>E_\lambda=\{\vec x \mid A\vec x=\lambda \vec x\}</m> is a subspace
of <m>\R^n</m>.
</p>
</statement>
<proof>
<p>
From <xref ref="SubspaceDefinition" /> it is sufficient
to show that two properties of closure under addition and
of closure under scalar multiplication are satisfied.
Suppose <m>\vec x</m> and <m>\vec y</m> are in <m>E_\lambda</m>.
<ul>
<li><p>
Closure under addition:
<md>
<mrow>A(\vec x+\vec y)\amp=A(\vec x)+A(\vec y)</mrow>
<mrow>\amp=\lambda \vec x + \lambda \vec y </mrow>
<mrow>\amp= \lambda (\vec x + \vec y) </mrow>
</md>
</p></li>
<li><p>
Closure under scalar multiplication:
<md>
<mrow>A(r\vec x)\amp=rA(\vec x)</mrow>
<mrow>\amp=r(\lambda \vec x)</mrow>
<mrow>\amp= \lambda (r\vec x)</mrow>
</md>
</p></li>
</ul>
</p>
</proof>
</proposition>
</section>
<section><title>Computing eigenvalues</title>
<p>
An easy but important fact is used to compute eigenvalues.
</p>
<proposition> <title>Eigenvalues <m>x</m> satisfy
<m>(A-\lambda I)\vec x=\vec0</m> </title>
<statement>
<p>
<me>
A\vec x=\lambda \vec x \textrm{ if and only if } (A-\lambda I)\vec x=\vec0
</me>
</p>
</statement>
<proof>
<p>
If <m>A\vec x=\lambda\vec x</m>, then
<me>
\vec0=A\vec x -\lambda\vec x=A\vec x -\lambda I\vec x
= (A - \lambda I)\vec x.
</me>
</p>
</proof>
</proposition>
<corollary>
<statement>
<p>
<m>\lambda</m> is an eigenvalue of <m>A</m> if and only if <m>A-\lambda I</m> is singular.
</p>
</statement>
<proof>
<p>
One of the equivalent conditions of singularity given in
<xref ref="InvertibilityEquivalence2" />
is that <m>B</m> is singular if <m>B\vec x=\vec 0</m> for some
<m>\vec x\not=\vec 0</m>. The matrix <m>A-\lambda I</m> plays the
role of <m>B</m>.
</p>
</proof>
</corollary>
<corollary>
<statement>
<p>
<m>\lambda</m> is an eigenvalue of <m>A</m> if and only if
<m>\det (A-\lambda I)=0</m>
</p>
</statement>
<proof>
<p>
<m>B</m> is singular if and only if <m>\det B=0</m>.
</p>
</proof>
</corollary>
<p>
We now can see how these corollaries allow us to compute eigenvalues.
We look at the matrix from <xref ref="EigenvalueExample1" />.
</p>
<example>
<statement>
<p>
We use
<me>
A-\lambda I=
\begin{bmatrix}
5-\lambda\amp -1\amp -2\\
1\amp 3-\lambda\amp -2\\
-1\amp -1\amp 4-\lambda
\end{bmatrix}
</me>
and
<me>
\det(A-\lambda I)=0.
</me>
Evaluating the determinant gives
<md>
<mrow>
\det\amp \begin{bmatrix}
5-\lambda\amp -1\amp -2\\
1\amp 3-\lambda\amp -2\\
-1\amp -1\amp 4-\lambda
\end{bmatrix}
</mrow>
<mrow>
\amp=(5-\lambda)(3-\lambda)(4-\lambda)-2+2
-2(5-\lambda) -2(3-\lambda)+(4-\lambda)
</mrow>
<mrow>\amp = -\lambda^3 +12\lambda^2 -44\lambda +48</mrow>
<mrow>\amp = -(\lambda-2)(\lambda-4)(\lambda-6)</mrow>
</md>
So we see that <m>\det(A-\lambda I)=0</m> only if
<m>\lambda=2</m>, <m>\lambda=4</m> or <m>\lambda=6</m>.
The evaluation of <m>\det(A-\lambda I)=0</m> gave us a
polynomial <m>p(\lambda)=-(\lambda-2)(\lambda-4)(\lambda-6)</m>
whose roots are the eigenvalues.
</p>
</statement>
</example>
<definition><title>The characteristic polynomial</title>
<statement>
<p>
The <term>characteristic polynomial</term> of a square matrix <m>A</m>
is the polynomial
<me>p(\lambda)=\det(\lambda I-A).</me>
</p>
</statement>
</definition>
<note>
<p>
Some sources define the characteristic polynomial as <m>\det(A-\lambda I)</m>.
Since <m>\lambda I-A=-(A-\lambda I)</m>, it follows that for any
square matrix <m>A</m> of size <m>n</m>,
<me>
\det(\lambda I-A)=(-1)^n\det(A-\lambda I)
</me>
and so both polynomials have the same roots.
</p>
</note>
<p>
We look at the matrix from <xref ref="EigenvalueExample2" />.
</p>
<example>
<statement>
<p>
<m>A=\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}</m>.
<me>
\begin{array}{rl}
\det(A-\lambda I)
\amp =\det\begin{bmatrix}2-\lambda\amp 1\amp 4\\
0\amp 3-\lambda\amp 0\\
2\amp -2\amp -5-\lambda \end{bmatrix}\\
\amp =(2-\lambda)(3-\lambda)(-5-\lambda)-8(3-\lambda)\\
\amp = (\lambda+6)(\lambda-3)^2
\end{array}
</me>
and so the eigenvalues are <m>\lambda=-6</m> and <m>\lambda=3</m>
</p>
</statement>
</example>
</section>
<section><title>Computing eigenspaces</title>
<p>
We have already defined eigenspaces in <xref ref="EigenspaceDef" />
</p>
<p>
Suppose we are given a square matrix <m>A</m> of order <m>n</m>,
a real number <m>\lambda</m>,
and we want to find all vectors in
<me>
E_\lambda=\{\vec x\in \R^n\mid A\vec x=\lambda\vec x\}
</me>.
If <m>A\vec x=\lambda\vec x</m>, then <m>A\vec x-\lambda\vec x=\vec 0</m>
and <m>(A-\lambda I)\vec x=\vec 0</m>. Hence we need only solve a system
of homogeneous equations.
</p>
<p>
From the previous <xref ref="EigenvalueExample1" />,
<me>
A=\begin{bmatrix}5\amp -1\amp -2\\ 1\amp 3\amp -2\\ -1\amp -1\amp 4 \end{bmatrix}
</me>
has eigenvalues <m>\lambda=2,4,6</m>. We find the eigenspaces for each
eigenvalue.
</p>
<example><title><m>\lambda=2</m></title>
<p>
<me>
A-2I=
\begin{bmatrix}5\amp -1\amp -2\\ 1\amp 3\amp -2\\ -1\amp -1\amp 4 \end{bmatrix}
- 2\begin{bmatrix}1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\end{bmatrix}
=\begin{bmatrix}3\amp -1\amp -2\\ 1\amp 1\amp -2\\ -1\amp -1\amp 2 \end{bmatrix}
</me>
As usual, we put the augmented matrix into reduced row echelon form:
<me>
\left[\begin{array}{ccc|c}
3\amp -1\amp -2\amp 0\\ 1\amp 1\amp -2\amp 0\\ -1\amp -1\amp 2\amp 0
\end{array}\right]
\textrm{ reduces to }
\left[\begin{array}{ccc|c}
1\amp 0\amp -1\amp 0\\ 0\amp 1\amp -1\amp 0\\ 0\amp 0\amp 0\amp 0
\end{array}\right]
</me>
and so all solutions are of the form <m>(x,y,z)=(t,t,t)=t(1,1,1)</m>.
</p>
</example>
<example><title><m>\lambda=4</m></title>
<p>
<me>
\left[\begin{array}{ccc|c}
1\amp -1\amp -2\amp 0\\
1\amp -1\amp -2\amp 0\\
-1\amp -1\amp 0\amp 0
\end{array}\right]
\textrm{ reduces to }
\left[\begin{array}{ccc|c}
1\amp 0\amp -1\amp 0\\
0\amp 1\amp 1\amp 0\\
0\amp 0\amp 0\amp 0
\end{array}\right]
</me>
and so all solutions are of the form <m>(x,y,z)=(t,-t,t)=t(1,-1,1)</m>.
</p>
</example>
<example><title><m>\lambda=6</m></title>
<p>
<me>
\left[\begin{array}{ccc|c}
-1\amp -1\amp -2\amp 0\\
1\amp -3\amp -2\amp 0\\
-1\amp -1\amp -2\amp 0
\end{array}\right]
\textrm{ reduces to }
\left[\begin{array}{ccc|c}
1\amp 0\amp 1\amp 0\\
0\amp 1\amp 1\amp 0\\
0\amp 0\amp 0\amp 0
\end{array}\right]
</me>
and so all solutions are of the form <m>(x,y,z)=(-t,-t,t)=t(-1,-1,1)</m>.
</p>
</example>
<p>
Notice that setting <m>t=1</m> in each case gives us the the original
eigenvectors of the example.
</p>
<p>
We can use similar arguments for
<xref ref="EigenvalueExample2" />, in which
<m>A=\begin{bmatrix}
2\amp 1\amp 4\\
0\amp 3\amp 0\\
2\amp -2\amp -5
\end{bmatrix}</m>
and <m>\lambda=-6,3</m>:
</p>
<example><title><m>\lambda=-6</m></title>
<p>
<me>
A-\lambda I=
\begin{bmatrix}8\amp 1\amp 4\\ 0\amp 9\amp 0\\ 2\amp -2\amp 1 \end{bmatrix}
\text{ reduces to }
\begin{bmatrix}1\amp 0\amp \frac12\\ 0\amp 1\amp 0\\ 0\amp 0\amp 0 \end{bmatrix}
</me>
and so all eigenvectors are of the form <m>(x,y,z)=t(1,0,-2)</m>.
</p>
</example>
<example><title><m>\lambda=3</m></title>
<p>
<me>
A-\lambda I=
\begin{bmatrix}-1\amp 1\amp 4\\ 0\amp 0\amp 0\\ 2\amp -2\amp -8 \end{bmatrix}
\text{ reduces to }
\begin{bmatrix}1\amp -1\amp -4\\ 0\amp 0\amp 0\\ 0\amp 0\amp 0 \end{bmatrix}
</me>
and so all eigenvectors are of the form <m>(x,y,z)=t(1,1,0)+u(4,0,1)</m>.
</p>
</example>
<proposition>
<title>Eigenvalues and the powers of a matrix</title>
<statement>
<p>
If <m>A\vec x=\lambda \vec x</m> then <m>A^n\vec x=\lambda^n \vec x</m>
for <m>n=1,2,\ldots</m>.
</p>
</statement>
<proof>
<p>
<md>
<mrow>
A^2\vec x=A(A\vec x)=A(\lambda \vec x)
=\lambda A\vec x=\lambda^2\vec x
</mrow>
<mrow>
A^3\vec x=A(A^2\vec x)=A(\lambda^2 \vec x)
=\lambda^2 A\vec x=\lambda^3\vec x
</mrow>
</md>
Repeating this process yields the desired result.
</p>
</proof>
</proposition>
<proposition>
<title>Eigenspaces and the powers of a matrix</title>
<statement>
<p>
Let <m>A</m> be an <m>n\times n</m> matrix having <m>\vec x_1,\ldots,\vec x_m</m>
as eigenvectors with <m>\lambda_1,\ldots,\lambda_m</m> as corresponding
eigenvalues. Further, let
<m>\vec x\in \Span\{\vec x_1,\ldots,\vec x_m\}</m>.
Then, for some <m>r_1,\ldots,r_n</m>,
<ul>
<li><p>
<m>A\vec x=\sum_{i=1}^m r_i\lambda_i\vec x_i</m>, and
</p></li>
<li><p>
<m>A^n\vec x=\sum_{i=1}^m r_i\lambda_i^n\vec x_i</m>
for <m>n\ge1</m>.
</p></li>
</ul>
</p>
</statement>
<proof>
<p>
By definition of the span of a set,
<me>
\vec x=\sum_{i=1}^m r_i\vec x_i
</me>
and so
<me>
A(\vec x)=A(\sum_{i=1}^m r_i\vec x_i)=\sum_{i=1}^m r_iA(\vec x_i)
=\sum_{i=1}^mr_i\lambda_i\vec x_i
</me>.
</p>
<p>
Similarly,
<me>
A^n(\vec x)=A^n(\sum_{i=1}^m r_i\vec x_i)=\sum_{i=1}^m r_iA^n(\vec x_i)
=\sum_{i=1}^mr_i\lambda_i^n\vec x_i
</me>.
</p>
</proof>
</proposition>
<example>
<title>Eigenspaces and the powers of a matrix</title>
<p>
Let
<m>A=\begin{bmatrix}0\amp1\amp1\\1\amp0\amp1\\1\amp1\amp0 \end{bmatrix}</m>,
<m>\vec x_1=\begin{bmatrix}1\\1\\1\end{bmatrix}</m>,
<m>\vec x_2=\begin{bmatrix}-1\\1\\0\end{bmatrix}</m>, and
<m>\vec x_3=\begin{bmatrix}-1\\0\\1\end{bmatrix}</m>.
Then <m>\vec x_1</m>, <m>\vec x_2</m>, and <m>\vec x_3</m>, are eigenvectors
of <m>A</m> with corresponding eigenvalues <m>2</m>, <m>-1</m>, and <m>-1</m>.
In fact <m>\{\vec x_1, \vec x_2, \vec x_3\}</m> is a basis for <m>\R^3</m>.
From the definition of a basis, for any given <m>\vec x\in\R^3</m>, there is
a unique choice of <m>r_1,r_2,r_3</m> so that
<m>\vec x=r_1\vec x_1 + r_2\vec x_2 + r_3\vec x_3</m>.
From the previous proposition,
<me>
A^n\vec x=2^nr_1\vec x_1+ (-1)^n r_2\vec x_2 +(-1)^n r_3 \vec x_3
</me>
As <m>n</m> gets large, the coefficient of <m>x_1</m> becomes huge and the
value of <m>A^n\vec x</m> is very close to a scalar multiple of <m>\vec x_1</m>,
that is, it approaches the eigenspace <m>E_2</m>.
</p>
</example>
</section>
<section><title>The number of eigenvalues of a matrix of order <m>n</m></title>
<p>
If <m>A</m> is a matrix of order <m>n</m>, then
<m>p(\lambda)=\det(A-\lambda I)</m> is a polynomial
of degree <m>n</m>. Since a polynomial of degree <m>n</m> has at most
<m>n</m> roots, the matrix has as most <m>n</m> eigenvalues.
</p>
<theorem xml:id="NumberOfEigenvalues">
<title>The number of eigenvalues of <m>A</m></title>
<statement>
<p>
A square matrix of order <m>n</m> has at most <m>n</m> eigenvalues.
</p>
</statement>
<proof>
<p>
The characteristic polynomial <m>p(\lambda)</m> is of degree <m>n</m>,
and such a polynomial has at most <m>n</m> real roots.
</p>
</proof>
</theorem>
<p>
If we once again look at the matrix from <xref ref="EigenvalueExample4" />.
we see that it is a matrix of order <m>4</m> with <m>4</m> eigenvalues given.
<xref ref="NumberOfEigenvalues" /> tells us that there are no others.
</p>
</section>
<section><title>Similarity and diagonalization</title>
<introduction>
<p>
Triangular matrices (including diagonal matrices in particular) have
eigenvalues that are particularly easy to compute. In fact, they are just
the diagonal entries.
</p>
<theorem xml:id="EigenvaluesTriangularMatrix">
<title>Eigenvalues of a triangular matrix </title>
<statement>
<p>
Let <m>A=[a_{i,j}]</m> be a triangular matrix of order <m>n</m>.
Then the eigenvalues of <m>A</m> are the diagonal entries
<m>a_{1,1},a_{2,2},\ldots,a_{n,n}</m>.
</p>
</statement>
<proof>
<p>
The matrix <m>A-\lambda I</m> is also triangular and, by
<xref ref="DeterminantTriangularMatrix" />,
its determinant is just the product of the diagonal elements, that is,
<me>
p(\lambda)=(a_{1,1}-\lambda)(a_{2,2}-\lambda)\cdots(a_{n,n}-\lambda).
</me>
The roots of this polynomial are the diagonal elements of <m>A</m>.
</p>
</proof>
</theorem>
<p>
Sometimes it is possible to find the eigenvalues of a matrix
by showing that it has the same eigenvalues as some diagonal matrix.
The key concept used to do this is
<term>similarity</term>.
</p>
</introduction>
<subsection>
<title>Similarity</title>
<definition>
<statement>
<p>
Let <m>A</m> and <m>B</m> be two square matrices of order <m>n</m>.
They are <term>similar</term> if there exists an invertible matrix
<m>P</m> of order <m>n</m> so that
<me>B=P^{-1}AP</me>.
</p>
</statement>
</definition>
<p>
The first goal is to show that similar matrices have identical eigenvalues.
</p>
<theorem xml:id="SimilarMatricesSameEigenvalues">
<title>Similar matrices have the same eigenvalues</title>
<statement>
<p>
If <m>A</m> and <m>B</m> are similar, then they have the
same eigenvalues.
</p>
</statement>
<proof>
<p>
Suppose <m>A\vec x=\lambda\vec x</m> with <m>\vec x\not=0</m>.
Let <m>\vec y=P^{-1}\vec x</m>.
Then
<me>
B\vec y = BP^{-1}\vec x
= P^{-1}APP^{-1}\vec x = P^{-1}A\vec x
= P^{-1}\lambda \vec x
= \lambda P^{-1}\vec x
= \lambda \vec y
</me>.
Note that <m>\vec y=\vec0</m> implies
<m>
\vec x = PP^{-1}\vec x =P\vec y = P\vec0 = \vec0
</m>
and, since <m>\vec x</m> is an eigenvector, this can not be the case.
Hence <m>\vec y\not=\vec0</m>, and
<m>\lambda</m> is an eigenvalue of <m>B</m>.
</p>
</proof>
</theorem>
<p>
If a matrix <m>A</m> is similar to a diagonal matrix, then
the eigenvalues are known.</p>
<theorem>
<statement>
<p>
If a matrix <m>A</m> is similar to a diagonal matrix <m>D</m>,
then the eigenvalues of <m>A</m> are the diagonal entries of <m>D</m>.
</p>
</statement>
<proof>
<p>
<xref ref="SimilarMatricesSameEigenvalues"/> implies that <m>A</m>
and <m>D</m> have the same eigenvalues, and
<xref ref="EigenvaluesTriangularMatrix"/> implies that the eigenvalues
of <m>D</m> are its diagonal elements.
</p>
</proof>
</theorem>
<p>
A matrix <m>A</m> being similar to a diagonal matrix is important:
it merits its own definition.
</p>
<definition>
<statement>
<p>
A matrix <m>A</m> is <term>diagonalizable</term> if it is similar
to a diagonal matrix.
</p>
</statement>
</definition>
<example xml:id="DiagonalizationExample">
<title>A diagonalizable matrix</title>
<p>
Let
<md>
<mrow>
A \amp =
\begin{bmatrix}
0 \amp 1 \amp 1 \amp 1 \\
1 \amp 0 \amp 1 \amp 1 \\
1 \amp 1 \amp 0 \amp 1 \\
1 \amp 1 \amp 1 \amp 0
\end{bmatrix}
\amp
\text{and}
\amp
\amp P \amp =
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1 \\
1 \amp -1 \amp 0 \amp 0 \\
1 \amp 0 \amp -1 \amp 0 \\
1 \amp 0 \amp 0 \amp -1
\end{bmatrix}
</mrow></md>
Then
<me>P^{-1}=\frac14
\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1 \\
1 \amp -3 \amp 1 \amp 1 \\
1 \amp 1 \amp -3 \amp 1 \\
1 \amp 1 \amp 1 \amp -3
\end{bmatrix}
</me>
and
<me>
P^{-1}AP=
\begin{bmatrix}
3 \amp 0 \amp 0\amp 0\\
0 \amp -1 \amp 0\amp 0\\
0 \amp 0 \amp -1\amp 0\\
0 \amp 0 \amp 0\amp -1
\end{bmatrix}
</me>.
Hence the eigenvalues of <m>A</m> are <m>3</m> and <m>-1</m>.
</p>
</example>
<p>
In <xref ref="DiagonalizationExample"/> the matrix <m>P</m>
is used to diagonalize a matrix. Where did it come from?
Under the right conditions,
eigenvectors may be used to construct this matrix.
</p>
<proposition xml:id="ConstructingPWithEigenvectors">
<title>Constructing <m>P</m> using eigenvectors</title>
<statement>
<p>
Suppose <m>A</m> is a square matrix of order <m>n</m>, and
<m>\{\vec x_1, \vec x_2,\ldots,\vec x_n\}</m> is a linearly
independent set of eigenvectors; say
<m>A\vec x_i=\lambda_i \vec x_i</m> for <m>i=1,2,\ldots,n</m>.
Let
<me>P=
\begin{bmatrix} \vec x_1\amp \vec x_2\amp\ldots\amp \vec x_n \end{bmatrix}
</me>,
that is, the eigenvectors are the columns of <m>P</m>.
Then
<me>
P^{-1}AP=\diag(\lambda_1,\lambda_2,\ldots,\lambda_n)
</me>.
</p>
</statement>
<proof>
<p>
From <xref ref="BasisMatrixNonSingular"/>, the matrix <m>P</m>
is nonsingular, and so <m>P^{-1}</m> exists. Now let
<m>\{\vec y_1, \vec y_2,\ldots,\vec y_n\}</m> be the rows of
<m>P^{-1}</m> so that
<me>
P^{-1}=
\begin{bmatrix} \vec y_1\\ \vec y_2\\ \vdots\\ \vec y_n \end{bmatrix}
</me>.
Then <m>P^{-1}P=I</m> is the same as
<men xml:id="IAsDotProduct">
\vec y_i \cdot \vec x_j=
\begin{cases}
1 \amp \text{if } i=j\\
0 \amp \text{otherwise.}
\end{cases}
</men>
Now consider <m>P^{-1}AP</m>.
<me>
P^{-1}AP
=\begin{bmatrix} \vec y_1\\ \vec y_2\\ \vdots\\ \vec y_n \end{bmatrix}A
\begin{bmatrix} \vec x_1\amp \vec x_2\amp\ldots\amp \vec x_n \end{bmatrix}
=\begin{bmatrix} \vec y_1\\ \vec y_2\\ \vdots\\ \vec y_n \end{bmatrix}
\begin{bmatrix} \lambda_1\vec x_1\amp \lambda_2\vec x_2\amp\ldots\amp\lambda_n
\vec x_n \end{bmatrix}
</me>.
Applying <xref ref="IAsDotProduct"/>,
<me>
P^{-1}AP=\diag(\lambda_1,\lambda_2,\ldots,\lambda_n)
</me>.
</p>
</proof>
</proposition>
<p>
<xref ref="ConstructingPWithEigenvectors"/>
motivates the search for linear independent eigenvectors.
</p>
<lemma xml:id="TwoEigenvectorsLinearlyIndependent">
<title>Two eigenvectors from different eigenvalues
are linearly independent</title>
<statement>
<p>Let <m>\vec x_1</m> and <m>\vec x_2</m> be two eigenvectors of <m>A</m>
corresponding to different eigenvalues. Then
<m>\{\vec x_1,\vec x_2\}</m> is linearly independent.
</p>
</statement>
<proof>
<p>
Say <m>A\vec x_1=\lambda_1\vec x_1</m> and
<m>A\vec x_2=\lambda_2\vec x_2</m>, and consider the equation
<m>r_1\vec x_1+r_2\vec x_2=\vec0</m>. Multiplying both sides of
the equation by the matrix <m>A</m> and multiplying both sides
by <m>\lambda_2</m> gives two new equations:
<md>
<mrow>r_1\lambda_1\vec x_1+r_2\lambda_2\vec x_2=\vec0</mrow>
<mrow>r_1\lambda_2\vec x_1+r_2\lambda_2\vec x_2=\vec0</mrow>
</md>
which implies
<me> r_1(\lambda_1-\lambda_2)\vec x_1=\vec0 </me>,
and, since <m>\lambda_1\not=\lambda_2</m>
and <m>\vec x_1\not=\vec0</m>, we have
<m>r_1=0</m>. Similarly, <m>r_2=0</m> and so
<m>\{\vec x_1,\vec x_2\}</m> is linearly independent.
</p>
</proof>
</lemma>
<proposition xml:id="EigenvectorsLinearlyIndependent">
<title>Eigenvectors with distinct eigenvalues form a linearly
independent set</title>
<statement>
<p>
Let <m>\vec x_1,\vec x_2,\ldots,\vec x_t</m> be eigenvectors of <m>A</m>