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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-37"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">5.3</span> <span class="title">Computing eigenspaces</span>
</h2>
<p id="p-1127">We have already defined eigenspaces in <a class="xref" data-knowl="./knowl/EigenspaceDef.html" title="Definition 5.1.7: Eigenspaces">Definition 5.1.7</a></p>
<p id="p-1128">Suppose we are given a square matrix \(A\) of order \(n\text{,}\) a real number \(\lambda\text{,}\) and we want to find all vectors in</p>
<div class="displaymath">
\begin{equation*}
E_\lambda=\{\vec x\in \R^n\mid A\vec x=\lambda\vec x\}\text{.}
\end{equation*}
</div>
<p class="continuation">If \(A\vec x=\lambda\vec x\text{,}\) then \(A\vec x-\lambda\vec x=\vec 0\) and \((A-\lambda I)\vec x=\vec 0\text{.}\) Hence we need only solve a system of homogeneous equations.</p>
<p id="p-1129">From the previous <a class="xref" data-knowl="./knowl/EigenvalueExample1.html" title="Example 5.1.3: Eigenvalues of \(A=
\begin{bmatrix}
5\amp -1\amp -2\\
1\amp 3\amp -2\\
-1\amp -1\amp 4
\end{bmatrix} \)">Example 5.1.3</a>,</p>
<div class="displaymath">
\begin{equation*}
A=\begin{bmatrix}5\amp -1\amp -2\\ 1\amp 3\amp -2\\ -1\amp -1\amp 4 \end{bmatrix}
\end{equation*}
</div>
<p class="continuation">has eigenvalues \(\lambda=2,4,6\text{.}\) We find the eigenspaces for each eigenvalue.</p>
<article class="example example-like" id="example-58"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-58"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.1</span><span class="period">.</span><span class="space"> </span><span class="title">\(\lambda=2\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-58"><article class="example example-like"><div class="displaymath" id="p-1130">
\begin{equation*}
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}
\end{equation*}
</div>
<p class="continuation">As usual, we put the augmented matrix into reduced row echelon form:</p>
<div class="displaymath">
\begin{equation*}
\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]
\end{equation*}
</div>
<p class="continuation">and so all solutions are of the form \((x,y,z)=(t,t,t)=t(1,1,1)\text{.}\)</p></article></div>
<article class="example example-like" id="example-59"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-59"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.2</span><span class="period">.</span><span class="space"> </span><span class="title">\(\lambda=4\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-59"><article class="example example-like"><div class="displaymath" id="p-1131">
\begin{equation*}
\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]
\end{equation*}
</div>
<p class="continuation">and so all solutions are of the form \((x,y,z)=(t,-t,t)=t(1,-1,1)\text{.}\)</p></article></div>
<article class="example example-like" id="example-60"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-60"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.3</span><span class="period">.</span><span class="space"> </span><span class="title">\(\lambda=6\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-60"><article class="example example-like"><div class="displaymath" id="p-1132">
\begin{equation*}
\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]
\end{equation*}
</div>
<p class="continuation">and so all solutions are of the form \((x,y,z)=(-t,-t,t)=t(-1,-1,1)\text{.}\)</p></article></div>
<p id="p-1133">Notice that setting \(t=1\) in each case gives us the the original eigenvectors of the example.</p>
<p id="p-1134">We can use similar arguments for <a class="xref" data-knowl="./knowl/EigenvalueExample2.html" title="Example 5.1.4: Eigenvalues of \(\begin{bmatrix}2\amp 1\amp 4\\ 0\amp 3\amp 0\\ 2\amp -2\amp -5 \end{bmatrix}\)">Example 5.1.4</a>, in which \(A=\begin{bmatrix}
2\amp 1\amp 4\\
0\amp 3\amp 0\\
2\amp -2\amp -5
\end{bmatrix}\) and \(\lambda=-6,3\text{:}\)</p>
<article class="example example-like" id="example-61"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-61"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.4</span><span class="period">.</span><span class="space"> </span><span class="title">\(\lambda=-6\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-61"><article class="example example-like"><div class="displaymath" id="p-1135">
\begin{equation*}
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}
\end{equation*}
</div>
<p class="continuation">and so all eigenvectors are of the form \((x,y,z)=t(1,0,-2)\text{.}\)</p></article></div>
<article class="example example-like" id="example-62"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-62"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.5</span><span class="period">.</span><span class="space"> </span><span class="title">\(\lambda=3\).</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-62"><article class="example example-like"><div class="displaymath" id="p-1136">
\begin{equation*}
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}
\end{equation*}
</div>
<p class="continuation">and so all eigenvectors are of the form \((x,y,z)=t(1,1,0)+u(4,0,1)\text{.}\)</p></article></div>
<article class="proposition theorem-like" id="proposition-14"><h6 class="heading">
<span class="type">Proposition</span><span class="space"> </span><span class="codenumber">5.3.6</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenvalues and the powers of a matrix.</span>
</h6>
<p id="p-1137">If \(A\vec x=\lambda \vec x\) then \(A^n\vec x=\lambda^n \vec x\) for \(n=1,2,\ldots\text{.}\)</p></article><article class="hiddenproof" id="proof-93"><a data-knowl="" class="id-ref proof-knowl original" data-refid="hk-proof-93"><h6 class="heading"><span class="type">Proof<span class="period">.</span></span></h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-proof-93"><article class="hiddenproof"><div class="displaymath" id="p-1138">
\begin{gather*}
A^2\vec x=A(A\vec x)=A(\lambda \vec x)
=\lambda A\vec x=\lambda^2\vec x\\
A^3\vec x=A(A^2\vec x)=A(\lambda^2 \vec x)
=\lambda^2 A\vec x=\lambda^3\vec x
\end{gather*}
</div>
<p class="continuation">Repeating this process yields the desired result.</p></article></div>
<article class="proposition theorem-like" id="proposition-15"><h6 class="heading">
<span class="type">Proposition</span><span class="space"> </span><span class="codenumber">5.3.7</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenspaces and the powers of a matrix.</span>
</h6>
<p id="p-1139">Let \(A\) be an \(n\times n\) matrix having \(\vec x_1,\ldots,\vec x_m\) as eigenvectors with \(\lambda_1,\ldots,\lambda_m\) as corresponding eigenvalues. Further, let \(\vec x\in \Span\{\vec x_1,\ldots,\vec x_m\}\text{.}\) Then, for some \(r_1,\ldots,r_n\text{,}\)</p>
<ul class="disc">
<li id="li-371"><p id="p-1140">\(A\vec x=\sum_{i=1}^m r_i\lambda_i\vec x_i\text{,}\) and</p></li>
<li id="li-372"><p id="p-1141">\(A^n\vec x=\sum_{i=1}^m r_i\lambda_i^n\vec x_i\) for \(n\ge1\text{.}\)</p></li>
</ul></article><article class="hiddenproof" id="proof-94"><a data-knowl="" class="id-ref proof-knowl original" data-refid="hk-proof-94"><h6 class="heading"><span class="type">Proof<span class="period">.</span></span></h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-proof-94"><article class="hiddenproof"><p id="p-1142">By definition of the span of a set,</p>
<div class="displaymath">
\begin{equation*}
\vec x=\sum_{i=1}^m r_i\vec x_i
\end{equation*}
</div>
<p class="continuation">and so</p>
<div class="displaymath">
\begin{equation*}
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\text{.}
\end{equation*}
</div>
<p id="p-1143">Similarly,</p>
<div class="displaymath">
\begin{equation*}
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\text{.}
\end{equation*}
</div></article></div>
<article class="example example-like" id="example-63"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-63"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">5.3.8</span><span class="period">.</span><span class="space"> </span><span class="title">Eigenspaces and the powers of a matrix.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-63"><article class="example example-like"><p id="p-1144">Let \(A=\begin{bmatrix}0\amp1\amp1\\1\amp0\amp1\\1\amp1\amp0 \end{bmatrix}\text{,}\) \(\vec x_1=\begin{bmatrix}1\\1\\1\end{bmatrix}\text{,}\) \(\vec x_2=\begin{bmatrix}-1\\1\\0\end{bmatrix}\text{,}\) and \(\vec x_3=\begin{bmatrix}-1\\0\\1\end{bmatrix}\text{.}\) Then \(\vec x_1\text{,}\) \(\vec x_2\text{,}\) and \(\vec x_3\text{,}\) are eigenvectors of \(A\) with corresponding eigenvalues \(2\text{,}\) \(-1\text{,}\) and \(-1\text{.}\) In fact \(\{\vec x_1, \vec x_2, \vec x_3\}\) is a basis for \(\R^3\text{.}\) From the definition of a basis, for any given \(\vec x\in\R^3\text{,}\) there is a unique choice of \(r_1,r_2,r_3\) so that \(\vec x=r_1\vec x_1 + r_2\vec x_2 + r_3\vec x_3\text{.}\) From the previous proposition,</p>
<div class="displaymath">
\begin{equation*}
A^n\vec x=2^nr_1\vec x_1+ (-1)^n r_2\vec x_2 +(-1)^n r_3 \vec x_3
\end{equation*}
</div>
<p class="continuation">As \(n\) gets large, the coefficient of \(x_1\) becomes huge and the value of \(A^n\vec x\) is very close to a scalar multiple of \(\vec x_1\text{,}\) that is, it approaches the eigenspace \(E_2\text{.}\)</p></article></div></section></div></main>
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