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<main class="main"><div id="content" class="pretext-content"><section xmlns:svg="http://www.w3.org/2000/svg" class="section" id="section-35"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">5.1</span> <span class="title">Eigenvalues and eigenvectors: preliminaries</span>
</h2>
<p id="p-1097">Eigenvalues and eigenvectors are defined for a square matrix \(A.\)</p>
<article class="definition definition-like" id="definition-63"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">5.1.1</span><span class="period">.</span><span class="space"> </span><span class="title">The eigenvalue of a matrix.</span>
</h6>
<p id="p-1098">A number \(\lambda\) is an <dfn class="terminology">eigenvalue</dfn> of a square matrix \(A\) if</p>
<div class="displaymath">
\begin{equation*}
A\vec x=\lambda\vec x
\end{equation*}
</div>
<p class="continuation">for some \(\vec x\not=\vec0\text{.}\)</p></article><article class="definition definition-like" id="definition-64"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">5.1.2</span><span class="period">.</span><span class="space"> </span><span class="title">The eigenvector of a matrix.</span>
</h6>
<p id="p-1099">If</p>
<div class="displaymath">
\begin{equation*}
A\vec x=\lambda\vec x
\end{equation*}
</div>
<p class="continuation">for some \(\vec x\not=\vec0\text{,}\) then \(\vec x\) is called an <dfn class="terminology">eigenvector</dfn> of \(A\) corresponding to the eigenvalue \(\lambda\text{.}\)</p></article><p id="p-1100">Notice that if \(\vec x=\vec 0\text{,}\) then \(A\vec x=\lambda\vec x\) is simply the equation \(\vec0=\vec0\) for any value of \(\lambda\text{.}\) This is not too interesting, and so we always have the restriction \(\vec x\not=\vec0\text{.}\)</p>
<article class="example example-like" id="EigenvalueExample1"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-EigenvalueExample1"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.1.3</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenvalues of \(A=
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix} \).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-EigenvalueExample1"><article class="example example-like"><div class="displaymath" id="p-1101">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes \(\vec x=\begin{bmatrix}1\\1\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=2\text{.}\)</p>
<div class="displaymath" id="p-1102">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes makes \(\vec
x=\begin{bmatrix}
1\\-1\\1
\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=4\text{.}\)</p>
<div class="displaymath" id="p-1103">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes makes \(\vec x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=6\text{.}\)</p>
<p id="p-1104">We have now computed eigenvalues of \(A\text{:}\) \(\lambda=2\text{,}\) \(\lambda=4\) and \(\lambda=6\text{.}\) With a little more theory, we will see that there are no others.</p></article></div>
<article class="example example-like" id="EigenvalueExample2"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-EigenvalueExample2"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.1.4</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenvalues of \(\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-EigenvalueExample2"><article class="example example-like"><p id="p-1105">Let \(A=\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\) Then we have</p>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes \(\vec x=\begin{bmatrix}1\\1\\0\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=3\text{.}\) Also</p>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes \(\vec x=\begin{bmatrix}4\\0\\1\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=3\) Finally</p>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">which makes \(\vec x=\begin{bmatrix}-1\\0\\2\end{bmatrix}\) an eigenvector with eigenvalue \(\lambda=-6\) Hence the demonstrated eigenvalues are \(\lambda=3\) and \(\lambda=-6\text{.}\) We will soon see that there are no more.</p></article></div>
<article class="example example-like" id="example-54"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-54"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.1.5</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenvalues of \(\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-54"><article class="example example-like"><p id="p-1106">Let \(A=\begin{bmatrix} 0\amp -1\\ 1\amp 0\end{bmatrix}\) and consider the equation \(A\vec x=\lambda\vec x\text{.}\) Setting \(\vec x=\begin{bmatrix} x_1\\x_2\end{bmatrix}\text{,}\)</p>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">This is equivalent to the system of equations</p>
<div class="displaymath">
\begin{equation*}
\begin{array}{rl}
\lambda x_1 \amp = -x_2\\
\lambda x_2 \amp = x_1.
\end{array}
\end{equation*}
</div>
<p class="continuation">This implies that \((\lambda^2+1)x_1 = \lambda^2 x_1 + x_1 = -\lambda x_2 +\lambda x_2=0 \) and \((\lambda^2+1)x_2 = \lambda^2 x_2 + x_2 = \lambda x_1 -\lambda x_21=0.\) Since \(\vec x \not=\vec 0\text{,}\) either \(x_1\not=0\) or \(x_2\not=0\text{,}\) and consequently \(\lambda^2=-1\text{.}\) Since no real number satisfies this equation, we conclude that there are no eigenvalues or eigenvectors for \(A\text{.}\) An aside: if we consider complex numbers, then \(i\) and \(-i\) are both eigenvalues of \(A\) with \(\begin{bmatrix} 1\\-i\end{bmatrix}\) and \(\begin{bmatrix} 1\\i\end{bmatrix}\) as corresponding eigenvectors.</p></article></div>
<article class="example example-like" id="EigenvalueExample4"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-EigenvalueExample4"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.1.6</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenvalues of \(\begin{bmatrix}
1 \amp 1 \amp 1 \amp 1\\
0 \amp 2 \amp 2 \amp 2\\
0 \amp 0 \amp 3 \amp 3\\
0 \amp 0 \amp 0 \amp 4
\end{bmatrix}\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-EigenvalueExample4"><article class="example example-like"><p id="p-1107">Consider the matrix</p>
<div class="displaymath">
\begin{equation*}
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}
\end{equation*}
</div>
<p class="continuation">Then it's easy to verify that</p>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<div class="displaymath">
\begin{equation*}
\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}
\end{equation*}
</div>
<p class="continuation">and so we have \(1\text{,}\) \(2\text{,}\) \(3\) and \(4\) 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 <a class="xref" data-knowl="./knowl/EigenvaluesTriangularMatrix.html" title="Theorem 5.5.1: Eigenvalues of a triangular matrix">Theorem 5.5.1</a> that for every upper (or lower) triangular matrix, the eigenvalues are the diagonal entries.</p></article></div>
<article class="definition definition-like" id="EigenspaceDef"><h6 class="heading">
<span class="type">Definition</span><span class="space"> </span><span class="codenumber">5.1.7</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenspaces.</span>
</h6>
<p id="p-1108">Suppose that \(A\) is a square matrix of order \(n\text{.}\) Then for any real number \(\lambda\text{,}\) we define the <dfn class="terminology">eigenspace</dfn> \(E_\lambda\) by</p>
<div class="displaymath">
\begin{equation*}
E_\lambda=\{\vec x\in \R^n\mid A\vec x=\lambda \vec x\}.
\end{equation*}
</div></article><p id="p-1109">Clearly \(\vec0\) is in \(E_\lambda\) for any value of \(\lambda\text{,}\) and \(\lambda\) is an eigenvalue if and only if there is some \(\vec x\not=\vec0\) in \(E_\lambda\text{.}\)</p>
<article class="proposition theorem-like" id="proposition-12"><h6 class="heading">
<span class="type">Proposition</span><span class="space"> </span><span class="codenumber">5.1.8</span><span class="period">.</span><span class="space"> </span><span class="title">An eigenspace is a subspace.</span>
</h6>
<p id="p-1110">Let \(A\) be an \(n\times n\) matrix and let \(\lambda\) be a real number. Then the set of vectors \(E_\lambda=\{\vec x \mid A\vec x=\lambda \vec x\}\) is a subspace of \(\R^n\text{.}\)</p></article><article class="hiddenproof" id="proof-89"><a data-knowl="" class="id-ref proof-knowl original" data-refid="hk-proof-89"><h6 class="heading"><span class="type">Proof<span class="period">.</span></span></h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-proof-89"><article class="hiddenproof"><p id="p-1111">From <a class="xref" data-knowl="./knowl/SubspaceDefinition.html" title="Definition 4.9.17: Subspace of \(\R^n\)">Definition 4.9.17</a> it is sufficient to show that two properties of closure under addition and of closure under scalar multiplication are satisfied. Suppose \(\vec x\) and \(\vec y\) are in \(E_\lambda\text{.}\)</p>
<ul class="disc">
<li id="li-369">
<p id="p-1112">Closure under addition:</p>
<div class="displaymath">
\begin{align*}
A(\vec x+\vec y)\amp=A(\vec x)+A(\vec y)\\
\amp=\lambda \vec x + \lambda \vec y \\
\amp= \lambda (\vec x + \vec y)
\end{align*}
</div>
</li>
<li id="li-370">
<p id="p-1113">Closure under scalar multiplication:</p>
<div class="displaymath">
\begin{align*}
A(r\vec x)\amp=rA(\vec x)\\
\amp=r(\lambda \vec x)\\
\amp= \lambda (r\vec x)
\end{align*}
</div>
</li>
</ul></article></div></section></div></main>
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