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<p>Die Matrix ist nicht positiv definit.</p>

<p>Berechne die Hauptminoren und prüfe ihr Vorzeichen:</p>


<p>\begin{align*}<br />\operatorname{det}\left(a_{11}\right) &amp;=\operatorname{det}(-1) \\</p>
<p>&amp;=-1 &lt;0</p>
<p>\end{align*}</p>


<p>\begin{align*}</p>
<p>\operatorname{det}\left(\begin{array}{ll}<br />a_{11} &amp; a_{12} \\<br />a_{21} &amp; a_{22}<br />\end{array}\right)&amp;=\operatorname{det}\left(\begin{array}{cc}<br />-1 &amp; 1 \\<br />1 &amp; -1<br />\end{array}\right) \\</p>
<p>&amp;=0\end{align*}</p>


<p>\begin{align*}<br />\operatorname{det}(A) &amp;=\operatorname{det}\left(\begin{array}{ccc}<br />-1 &amp; 1 &amp; \sqrt{2} \\<br />1 &amp; -1 &amp; \sqrt{2} \\<br />\sqrt{2} &amp; \sqrt{2} &amp; 0<br />\end{array}\right) \\</p>
<p>&amp;=0+2+2-(-2-2+0) \\<br />&amp;=8 &gt;0<br />\end{align*}</p>


<p><mjx-container class="MathJax" jax="SVG"><svg alt="\begin{align*}
\operatorname{det}(A) &=\operatorname{det}\left(\begin{array}{ccc}
-1 & 1 & \sqrt{2} \\
1 & -1 & \sqrt{2} \\
\sqrt{2} & \sqrt{2} & 0
\end{array}\right) \\
&=0+2+2-(-2-2+0) \\
&=8 >0
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<p>Da nicht alle Hauptminoren positiv sind, ist die Matrix nicht positiv definit.</p>

<p>Berechne alle Hauptminoren und entscheide anschließend ob die Matrix positiv definit ist.</p>
<p> </p>
<p>$$A=\begin{align*}\left(\begin{array}{ccc}<br />-1 &amp; 1 &amp; \sqrt{2} \\<br />1 &amp; -1 &amp; \sqrt{2} \\<br />\sqrt{2} &amp; \sqrt{2} &amp; 0<br />\end{array}\right)\end{align*}$$</p>

<p>Berechne alle Hauptminoren und entscheide anschließend ob die Matrix positiv definit ist.</p>
<p> </p>
<p><mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt="A=\begin{align*}\left(\begin{array}{ccc}
-1 & 1 & \sqrt{2} \\
1 & -1 & \sqrt{2} \\
\sqrt{2} & \sqrt{2} & 0
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Dies ist eine interaktive Aufgabe zu: Definitheit prüfen mit praktischen Tipps zum Lösen und einer Zusammenfassung der nötigen Theorie

Eigenwerte - Definitheit prüfen | Aufgabe mit Lösung


<h2 class="content_sub_heading">Eigenwerten, Räume und Vektoren</h2>
<ol>
<li>Bestimme das charakteristische Polynom von $A$ über $\chi_{A}(\lambda)=\operatorname{det}(A-\lambda E_n)$. Ziehe dazu jeweils ein $\lambda$ von den Einträgen der Hauptdiagonalen ab und berechne anschließend die Determinante.</li>
<li>Die Eigenwerte entsprechen den Nullstellen des charakteristischen Polynoms.<br />$\chi_{A}(\lambda)=0 \quad \Rightarrow \quad \lambda_1, \ldots, \lambda_n$</li>
<li>Bestimme jeweils den Eigenraum zu jedem gefundenen Eigenwert mit $\operatorname{Eig}_{A}(\lambda_i)=\operatorname{ker}(A-\lambda_i \cdot E_n)$<br /><em>Du solltest hier bei jeder Berechnung unendlich viele Lösungen erhalten, also immer eine Lösungsmenge mit mindestens einem Parameter.</em></li>
<li>Einen passenden Eigenvektor erhälst du nun, indem du jeweils einen Vektor aus dem zugehörigen Eigenraum auswählst. Der Nullvektor ist dabei allerdings nicht zulässig.</li>
</ol>


Aufgabenstellung:

Lösungsweg:

Lösung: