The algebraic multiplicity of a root is the number of times it appears in the factorisation \(P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n)\).
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The algebraic multiplicity of a root is the number of times it appears in the factorisation \(P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n)\).
Example: If the algebraic multiplicity of \(\lambda_2\) is \(3\) then \((z - \lambda_2)^3 \ | \ P(z)\).
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| Text | The {{c1::algebraic multiplicity of a <i>root</i>}} is {{c2:: the number of times it appears in the factorisation \(P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n)\)}}. | |
| Extra | <div><strong>Example:</strong> If the algebraic multiplicity of \(\lambda_2\) is \(3\) then \((z - \lambda_2)^3 \ | \ P(z)\).</div> |
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\(\overline{z_1 + z_2} = {{c1:: \overline{z_1} + \overline{z_2} }}\)
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\(\overline{z_1 + z_2} = {{c1:: \overline{z_1} + \overline{z_2} }}\)
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| Text | \(\overline{z_1 + z_2} = {{c1:: \overline{z_1} + \overline{z_2} }}\) |
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Prove some \(x \in \mathbb{C}\) is actually in \(\mathbb{R}\)?
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Prove some \(x \in \mathbb{C}\) is actually in \(\mathbb{R}\)?
Show that \(x = \overline{x} \implies x \in \mathbb{R}\)
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| Front | <div>Prove some \(x \in \mathbb{C}\) is actually in \(\mathbb{R}\)?</div> | |
| Back | Show that \(x = \overline{x} \implies x \in \mathbb{R}\) |
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Every polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has {{c1:: a zero \(\lambda \in \mathbb{C} \)}}
Back
Every polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has {{c1:: a zero \(\lambda \in \mathbb{C} \)}}
Fundamental theorem of algebra
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| Text | Every polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has {{c1:: a zero \(\lambda \in \mathbb{C} \)}} | |
| Extra | Fundamental theorem of algebra |
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Express the Fibonacci formula \(f_n = f_{n - 1} + f_{n - 2}\) as a matrix equation?
- {{c1::Let \(\mathbf{g}_{n+1} = M\mathbf{g}_n\), where \(M = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)(matrix version of the recursion)}}
- The eigenvalues \(\lambda_1 = \frac{1+\sqrt{5{2}\) (golden ratio \(\phi\)) and \(\lambda_2 = \frac{1-\sqrt{5}}{2}\) are found, along with their eigenvectors \(v_1 = \begin{pmatrix} \lambda_1 \\ 1 \end{pmatrix}\) and \(v_2 = \begin{pmatrix} \lambda_2 \\ 1 \end{pmatrix}\) . These eigenvectors are independent since \(\lambda_1 \neq \lambda_2\) and thus they form a basis for \(\mathbb{R}^2\).}}
- {{c3::The initial state \(\mathbf{g}_0\) is written as a linear combination of eigenvectors with coefficients \(\pm\frac{1}{\sqrt{5}}\): \(g_0 = \frac{1}{\sqrt{5}}v_1 - \frac{1}{\sqrt{5}}v_2\).}}
- {{c4::Since \(M^n\mathbf{v}_i = \lambda_i^n\mathbf{v}_i\), we get the closed form:\[F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]\]}}
Back
Express the Fibonacci formula \(f_n = f_{n - 1} + f_{n - 2}\) as a matrix equation?
- {{c1::Let \(\mathbf{g}_{n+1} = M\mathbf{g}_n\), where \(M = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)(matrix version of the recursion)}}
- The eigenvalues \(\lambda_1 = \frac{1+\sqrt{5{2}\) (golden ratio \(\phi\)) and \(\lambda_2 = \frac{1-\sqrt{5}}{2}\) are found, along with their eigenvectors \(v_1 = \begin{pmatrix} \lambda_1 \\ 1 \end{pmatrix}\) and \(v_2 = \begin{pmatrix} \lambda_2 \\ 1 \end{pmatrix}\) . These eigenvectors are independent since \(\lambda_1 \neq \lambda_2\) and thus they form a basis for \(\mathbb{R}^2\).}}
- {{c3::The initial state \(\mathbf{g}_0\) is written as a linear combination of eigenvectors with coefficients \(\pm\frac{1}{\sqrt{5}}\): \(g_0 = \frac{1}{\sqrt{5}}v_1 - \frac{1}{\sqrt{5}}v_2\).}}
- {{c4::Since \(M^n\mathbf{v}_i = \lambda_i^n\mathbf{v}_i\), we get the closed form:\[F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]\]}}
Field-by-field Comparison
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| Text | Express the Fibonacci formula \(f_n = f_{n - 1} + f_{n - 2}\) as a matrix equation?<br><ol><li>{{c1::Let \(\mathbf{g}_{n+1} = M\mathbf{g}_n\), where \(M = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)(matrix version of the recursion)}}</li><li>{{c2::The eigenvalues \(\lambda_1 = \frac{1+\sqrt{5}}{2}\) (golden ratio \(\phi\)) and \(\lambda_2 = \frac{1-\sqrt{5}}{2}\) are found, along with their eigenvectors \(v_1 = \begin{pmatrix} \lambda_1 \\ 1 \end{pmatrix}\) and \(v_2 = \begin{pmatrix} \lambda_2 \\ 1 \end{pmatrix}\) . These eigenvectors are independent since \(\lambda_1 \neq \lambda_2\) and thus they form a basis for \(\mathbb{R}^2\).}}</li><li>{{c3::The initial state \(\mathbf{g}_0\) is written as a linear combination of eigenvectors with coefficients \(\pm\frac{1}{\sqrt{5}}\): \(g_0 = \frac{1}{\sqrt{5}}v_1 - \frac{1}{\sqrt{5}}v_2\).}}</li><li>{{c4::Since \(M^n\mathbf{v}_i = \lambda_i^n\mathbf{v}_i\), we get the closed form:\[F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]\]}}</li></ol> |
Note 6: ETH::LinAlg
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If you have some expression \(x_1 + i x_2 = y_1 + i y_2\) we can separate this into?
Back
If you have some expression \(x_1 + i x_2 = y_1 + i y_2\) we can separate this into?
- \(x_1 = y_1\)
- \(x_2 = y_2\)
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| Front | If you have some expression \(x_1 + i x_2 = y_1 + i y_2\) we can separate this into? | |
| Back | <ol><li>\(x_1 = y_1\)</li><li>\(x_2 = y_2\)</li></ol>because of the \(i\)<br> |
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Fundamental Theorem of Algebra?
Back
Fundamental Theorem of Algebra?
Any degree \(n\) non-constant \(n \geq 1\) polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has a (existance) zero: \(\lambda \in \mathbb{C}\) such that \(P(\lambda) = 0\).
Field-by-field Comparison
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| Front | Fundamental Theorem of Algebra? | |
| Back | <div>Any degree \(n\) non-constant \(n \geq 1\) polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has a (existance) zero: \(\lambda \in \mathbb{C}\) such that \(P(\lambda) = 0\).</div> |
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\(\frac{1}{z} = {{c1:: \frac{\overline{z} }{|z|^2} :: \text{in terms of z} }}\)
Back
\(\frac{1}{z} = {{c1:: \frac{\overline{z} }{|z|^2} :: \text{in terms of z} }}\)
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| Text | \(\frac{1}{z} = {{c1:: \frac{\overline{z} }{|z|^2} :: \text{in terms of z} }}\) |
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Was ist eine konjugiert-transponierte (auch: Hermitesch-transponierte) Matrix?
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Was ist eine konjugiert-transponierte (auch: Hermitesch-transponierte) Matrix?
\( \mathbf{A}^H = (\overline{\mathbf{A}})^\top = \overline{\mathbf{A}^\top}\)
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Was ist eine konjugiert-transponierte (auch: Hermitesch-transponierte) Matrix?
Back
Was ist eine konjugiert-transponierte (auch: Hermitesch-transponierte) Matrix?
\( \mathbf{A}^* = (\overline{\mathbf{A}})^\top = \overline{\mathbf{A}^\top}\)
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| Back | \( \mathbf{A}^ |
\( \mathbf{A}^* = (\overline{\mathbf{A}})^\top = \overline{\mathbf{A}^\top}\)<br> |
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Wann ist eine Matrix hermitesch?
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Wann ist eine Matrix hermitesch?
Falls \( \mathbf{A}^H = A\)
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Wann ist eine Matrix hermitesch?
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Wann ist eine Matrix hermitesch?
Falls \( \mathbf{A}^* = A\)
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| Back | Falls \( \mathbf{A}^ |
Falls \( \mathbf{A}^* = A\) |
Note 11: ETH::LinAlg
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Any degree \(n\) non constant (\(n \geq 1\)) polynomial \(P(z)\) has {{c1:: \(n\) zeros: \(\lambda_1, \dots, \lambda_n \in \mathbb{C}\)}}, perhaps with repetitions such that \[ P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n) \]
Back
Any degree \(n\) non constant (\(n \geq 1\)) polynomial \(P(z)\) has {{c1:: \(n\) zeros: \(\lambda_1, \dots, \lambda_n \in \mathbb{C}\)}}, perhaps with repetitions such that \[ P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n) \]
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| Text | Any degree \(n\) non constant (\(n \geq 1\)) polynomial \(P(z)\) has {{c1:: \(n\) zeros: \(\lambda_1, \dots, \lambda_n \in \mathbb{C}\)}}, perhaps {{c1:: with repetitions}} such that \[ P(z) = a_n (z-\lambda_1)(z - \lambda_2) \dots (z - \lambda_n) \]<br> |
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For a complex vector \(v\) we have \(||v|| =\) {{c1:: \(v^*v = \overline{v}^\top v = \sum_{i = 1}^n \overline{v_i}v_i = \sum_{i = 1}^n |v_i|^2\)}}.
Back
For a complex vector \(v\) we have \(||v|| =\) {{c1:: \(v^*v = \overline{v}^\top v = \sum_{i = 1}^n \overline{v_i}v_i = \sum_{i = 1}^n |v_i|^2\)}}.
Field-by-field Comparison
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| Text | For a complex vector \(v\) we have \(||v|| =\) {{c1:: \(v^*v = \overline{v}^\top v = \sum_{i = 1}^n \overline{v_i}v_i = \sum_{i = 1}^n |v_i|^2\)}}. |
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\(z\overline{z} = |z|^2 \)
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\(z\overline{z} = |z|^2 \)
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| Text | \(z\overline{z} = {{c1:: |z|^2 }} \) |