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      <text>22.3.2.5 Real symmetric matrices have real eigenvalues</text>
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      <text><![CDATA[<p>In particular any \(2\times 2\) matrix of the form
\[ A=\left[\begin{array}{rc}a&amp;b\\ -b&amp;a\end{array}\right]\]
has complex conjugate eigenvalues \(a\pm \mathrm{i} b\).</p>

<p>A \(3\times 3\) example of a matrix with some complex eigenvalues is
\[ B=\left[\begin{array}{crr}1&amp;-1&amp;-1\\ 1&amp;-1&amp;0\\ 1&amp;0&amp;-1\end{array}\right].\]
A straightforward calculation shows that the eigenvalues of \(B\) are
\[ \lambda=-1 \text{ (real)}, \lambda =\pm \mathrm{i} \text{ (complex conjugates)}.\]</p>

<p>With <strong>symmetric</strong> matrices on the other hand, complex eigenvalues are not possible.</p>
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>The eigenvalues of a symmetric matrix with real elements are always real.</p>
</div>

<p>The general proof of this result in the above Key Point is beyond our scope, but a simple proof for symmetric \(2\times 2\) matrices is straightforward.</p>
<p>Let \(A=\left[\begin{array}{cc}a&amp;b\\ b&amp;c\end{array}\right]\) be any \(2\times 2\) symmetric matrix, \(a,\ b,\ c\) being real numbers.</p>
<p>The characteristic equation for \(A\) is
\[ (a-\lambda)(c-\lambda)-b^2=0\]
or, expanding: \[ \lambda^2-(a+c)\lambda+ac-b^2=0 \]
from which
\[ \lambda=\dfrac{(a+c)\pm\sqrt{(a+c)^2-4ac+4b^2}}{2}.\]
The quantity under the square root sign can be treated as follows:
\[ (a+c)^2-4ac+4b^2=a^2+c^2+2ac-4ac+b^2=(a-c)^2+4b^2\]
which is always positive and hence \(\lambda\) <strong>cannot be complex</strong>.</p>]]></text>
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