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      <text>22.1.1.5 A simple eigenvalue problem</text>
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      <text><![CDATA[<p></p><h3>A simple eigenvalue problem</h3><p>We shall be interested in simultaneous equations of the form:</p><p>\(\qquad A\underline{x}=\lambda \underline{x},\)</p><p>where&nbsp;\(A\)&nbsp;is an&nbsp;\(n\times n\)&nbsp;matrix,&nbsp;\(\underline{x}\)&nbsp;is an&nbsp;\(n\times 1\)&nbsp;column vector and&nbsp;\(\lambda\)&nbsp;is a scalar (a constant) and, in the first instance, we examine some simple examples to gain experience of solving problems of this type.</p>
<p>We approach the following example by examining the equations. This should make the actions within the solution more transparent. In practice, we will work with the matrices directly which will give a much less cluttered solution.</p>

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<h4 class="HELM_example">Example</h4>
<p>Consider the following system with \(n=2\):</p>
\[\left\{\begin{alignedat}{2}
2x &amp; {}+{} &amp; 3y &amp; = \lambda x \\
3x &amp; {}+{} &amp; 2y &amp; = \lambda y
\end{alignedat}\right.\]
<p>so that</p>
<p style="text-align: center;">\(A=\begin{bmatrix}2&amp;3\\ 3&amp;2\end{bmatrix}\) &emsp; and &emsp; \(\underline{x}=\begin{bmatrix}x\\ y\end{bmatrix}\).</p>
<p>It appears that there are three unknowns&nbsp;\(x,y,\lambda\). The obvious questions to ask are: can we find&nbsp;\(x,y\)? what is&nbsp;\(\lambda\)?</p>
<h4 class="HELM_solution">Solution</h4>
<p>To solve this problem we firstly re-arrange the equations (take all unknowns onto one side);</p>
\begin{align}
(2-\lambda)x + 3y &amp;= 0 \tag{1} \\
3x + (2-\lambda)y &amp;= 0 \tag{2}
\end{align}
<p>Therefore, from equation \((2)\):</p>
\[x = -\frac{(2-\lambda)}{3}y \tag{3}\]
<p>Then when we substitute this into \((1)\)</p>
<p style="text-align:center;">\(\displaystyle -\frac{(2-\lambda)^2}{3}y+3y = 0\) &emsp; which simplifies to \([-(2-\lambda)^2+9]y=0\).</p>
<p>We conclude that either \(y=0\) or \(9=(2-\lambda)^2\). There are thus two cases to consider:</p>
<h5>Case 1</h5>
<p>If \(y=0\) then \(x=0\) (from \((3)\)) and we get the <b>trivial solution</b>. (We could have guessed this solution at the outset.)</p>
<h5>Case 2</h5>
<p>\(\qquad 9=(2-\lambda)^2\)</p>
<p>which gives, on taking square roots:</p>
<p style="text-align: center;">\(\pm 3 = 2-\lambda\) &emsp; giving &emsp; \(\lambda=2\pm 3\), &emsp; so &emsp; \(\lambda = 5\) &emsp; or &emsp; \(\lambda =1\).</p>
<p>Now, from equation \((3)\), if \(\lambda=5\) then \(x=+y\) and if \(\lambda=-1\) then \(x=-y\).</p>

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<p>We have now completed the analysis. We have found values for \(\lambda\) but we also see that we cannot obtain unique values for \(x\) and \(y\): all we can find is the ratio between these quantities. This behaviour is typical, as we shall now see, of an eigenvalue problem.</p>]]></text>
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