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      <text>22.1.3.1 Properties of eigenvalues and eigenvectors</text>
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      <text><![CDATA[<h3>Properties of eigenvalues and eigenvectors</h3><p>There are a number of general properties of eigenvalues and eigenvectors which you should be familiar with. You will be able to use them as a check on some of your calculations.&nbsp;</p><p><span><strong>Property 1: Sum of eigenvalues</strong></span></p><p>For any square matrix <span class="math inline">\(A\)</span>:&nbsp;</p><p>\(\quad\)sum of eigenvalues <span class="math inline">\(=\)</span> sum of diagonal terms of <span class="math inline">\(A\)</span> (called the <span><strong>trace</strong></span> of <span class="math inline">\(A\)</span>)</p><p>Formally, for an <span class="math inline">\(n\times n\)</span> matrix <span class="math inline">\(A\)</span>: <span class="math inline">\(\qquad \displaystyle{\sum_{i=1}^n \lambda_i=\ \text{ trace} (A)}\)</span></p><p>(Repeated eigenvalues must be counted according to their multiplicity.)<br>
Thus if <span class="math inline">\(\lambda_1=4, \lambda_2=4,
\lambda_3=1\)</span> then <span class="math inline">\(\displaystyle{\sum_{i=1}^{3}\lambda_i=9}\)</span>).</p><p><span><strong>Property 2: Product of eigenvalues</strong></span></p><p>For any square matrix <span class="math inline">\(A\)</span>:&nbsp;</p><p>\(\quad\)product of eigenvalues <span class="math inline">\(=\)</span> determinant of <span class="math inline">\(A\)</span></p><p>Formally: <span class="math inline">\(\displaystyle{\qquad \lambda_1\lambda_2\lambda_3\cdots\lambda_n=\prod_{i=1}^n \lambda_i=\text{ det}(A)}\)</span></p><p>The symbol <span class="math inline">\(\prod\)</span> simply denotes multiplication, as <span class="math inline">\(\sum\)</span> denotes summation.</p><hr>
<h4 class="HELM_example">Example<br></h4>
<p>Verify Properties 1 and 2 for the <span class="math inline">\(3\times 3\)</span> matrix:</p><p><span class="math inline">\(\qquad A = \left[\begin{array}{rrr}2&amp;-1&amp;0\\ -1&amp;2&amp;-1\\ 0&amp;-1&amp;2\end{array}\right]\)</span></p><p>whose eigenvalues were found earlier.</p>
<h4 class="HELM_solution">Solution</h4><p>The three eigenvalues for this matrix are:</p><p>\(\qquad \lambda_1=2-\sqrt{2},\qquad \lambda_2=2,\qquad \lambda_3=2+\sqrt{2}.\)</p><p>Therefore</p><p>\(\qquad \lambda_1+\lambda_2+\lambda_3=(2-\sqrt{2})+2+(2+\sqrt{2})=6=\text{trace}(A),\)</p><p>whilst</p><p>\(\qquad \lambda_1\lambda_2\lambda_3=(2-\sqrt{2})(2)(2+\sqrt{2})=4=\text{det}(A).\)</p><hr>
<p><strong>Property 3: Linear independence of eigenvectors</strong><br></p>
<p>Eigenvectors of a matrix <span class="math inline">\(A\)</span> corresponding to distinct eigenvalues are <span><strong>linearly independent</strong></span> i.e. one eigenvector <span><strong>cannot</strong></span> be written as a linear sum of the other eigenvectors. The proof of this result is omitted but we illustrate this property with two examples.</p><p>We saw earlier that the matrix&nbsp;</p><p><span class="math inline">\(\qquad A=\left[\begin{array}{cc}1&amp;0\\ 1&amp;2\end{array}\right]\)</span></p><p>has distinct eigenvalues <span class="math inline">\(\lambda_1=1\qquad \lambda_2=2\)</span> with associated eigenvectors</p><p><span class="math inline">\(\qquad \underline{x_1}=\dfrac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]\qquad
\underline{x_2}=\left[\begin{array}{r}0\\ 1\end{array}\right]\)</span></p><p>respectively.&nbsp;</p><p>Clearly <span class="math inline">\(\underline{x_1}\)</span> is <span><strong>not</strong></span> a constant multiple of <span class="math inline">\(\underline{x_2}\)</span> and these eigenvectors are <span><strong>linearly independent</strong></span>.&nbsp;</p><p>We also saw that the <span class="math inline">\(3\times 3\)</span> matrix&nbsp;</p><p><span class="math inline">\(\qquad A=\left[\begin{array}{rrr}2&amp;-1&amp;0\\ -1&amp;2&amp;-1\\ 0&amp;-1&amp;2\end{array}\right]\)</span></p><p>had the following distinct eigenvalues&nbsp;&nbsp;<span class="math inline">\(\lambda_1=2-\sqrt{2},\ \lambda_2=2,\ \lambda_3=2+\sqrt{2}\)</span>&nbsp; with corresponding eigenvectors of the form shown: <span class="math inline">\(\qquad \underline{x_1}=\left[\begin{array}{c}1\\ \sqrt{2}\\ 1\end{array}\right],\qquad
\underline{x_2}=\left[\begin{array}{r}1\\0\\-1\end{array}\right],\qquad
 \underline{x_3}=\left[\begin{array}{c}1\\ -\sqrt{2}\\ 1\end{array}\right]\)</span></p><p>Clearly none of these eigenvectors is a constant multiple of any other. Nor is any one obtainable as a linear combination of the other two. The three eigenvectors are <span><strong>linearly independent</strong></span>.</p><p><span><strong>Property 4: Eigenvalues of diagonal matrices</strong></span></p><p>A <span class="math inline">\(2\times 2\)</span> diagonal matrix <span class="math inline">\(D\)</span> has the form</p><p><span class="math inline">\(\qquad D=\left[\begin{array}{cc}a&amp;0\\ 0&amp;d\end{array}\right]\)</span></p><p>The characteristic equation</p><p><span class="math inline">\(\quad |D-\lambda I|=0\)&nbsp;</span>is <span class="math inline">\(\left|\begin{array}{cc}a-\lambda&amp;0\\ 0&amp;d-\lambda\end{array}\right|=0\)</span>&nbsp;</p><p>i.e. <span class="math inline">\(\qquad (a-\lambda)(d-\lambda)=0\)</span></p><p>So the eigenvalues are simply the diagonal elements <span class="math inline">\(a\)</span> and <span class="math inline">\(d\)</span>.</p><p>Similarly a <span class="math inline">\(3\times 3\)</span> diagonal matrix has the form</p><p><span class="math inline">\(\qquad D=\left[\begin{array}{ccc}a&amp;0&amp;0\\ 0&amp;b&amp;0\\ 0&amp;0&amp;c\end{array}\right]\)</span></p><p>having characteristic equation</p><p><span class="math inline">\(\qquad |D-\lambda I|=(a-\lambda)(b-\lambda)(c-\lambda)=0\)</span></p><p>so again the diagonal elements <span><strong>are</strong></span> the eigenvalues.</p><p>We can see that a diagonal matrix is a particularly simple matrix to work with. In addition to the eigenvalues being obtainable immediately by inspection it is exceptionally easy to multiply diagonal matrices.</p>]]></text>
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