Review Note
Last Update: 06/26/2024 10:58 AM
Current Deck: LA 2
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Duale Basis
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Sei \(V = \mathbb{R}^n\) dann gilt \(E=(e_1,\dots,e_n)\) ist Basis von V und
\(E \rightsquigarrow E^*=(e_1^*,\dots,e_n^*)\) da \(e_1= \left( \begin{matrix} 1 \\ 0 \\\vdots\\0 \end{matrix} \right) \Rightarrow e_1^* \left(\begin{matrix} x_1 \\\vdots\\x_n \end{matrix} \right)= \left( \begin{matrix} 1 \\ 0 \\\vdots\\0 \end{matrix} \right)\cdot\left( \begin{matrix} x_1 \\\vdots\\x_n \end{matrix} \right) = x_1\)
\(E \rightsquigarrow E^*=(e_1^*,\dots,e_n^*)\) da \(e_1= \left( \begin{matrix} 1 \\ 0 \\\vdots\\0 \end{matrix} \right) \Rightarrow e_1^* \left(\begin{matrix} x_1 \\\vdots\\x_n \end{matrix} \right)= \left( \begin{matrix} 1 \\ 0 \\\vdots\\0 \end{matrix} \right)\cdot\left( \begin{matrix} x_1 \\\vdots\\x_n \end{matrix} \right) = x_1\)
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