User:Sintel/Dual Weil-Euclidean norm: Difference between revisions
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→Relation to other metrics: -stray subscript |
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:<math> | :<math> | ||
W = \begin{bmatrix} | W = | ||
\log_2 2 & 0 & \cdots & 0 \\ | \begin{bmatrix} | ||
0 & \log_2 3 & \cdots & 0\\ | \log_2 2 & 0 & \cdots & 0 \\ | ||
\vdots & \vdots & \ddots & \vdots \\ | 0 & \log_2 3 & \cdots & 0\\ | ||
0 & 0 & \cdots & \log_2 p | \vdots & \vdots & \ddots & \vdots \\ | ||
\end{bmatrix} | 0 & 0 & \cdots & \log_2 p | ||
\end{bmatrix} | |||
</math> | </math> | ||
| Line 19: | Line 20: | ||
:<math> | :<math> | ||
X = \begin{bmatrix} | X = | ||
W \\ | \begin{bmatrix} | ||
\hline | W \\ | ||
j | \hline | ||
\end{bmatrix} | j | ||
\end{bmatrix} | |||
</math> | </math> | ||
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:<math> | :<math> | ||
\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} | \left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} | ||
||\alpha||^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}} | </math> | ||
:<math> | |||
||\alpha||^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}} | |||
</math> | </math> | ||
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:<math> | :<math> | ||
G = X^{\mathsf T} X = W^2 + j^{\mathsf T}j | G = X^{\mathsf T} X = W^2 + j^{\mathsf T}j | ||
</math> | </math> | ||
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<ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref> | <ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref> | ||
: If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and | ::If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and | ||
: <math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math> | ::<math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math> | ||
Now identifying <math>A = W^2</math> and <math>B = j^{\mathsf T}j</math>. We can see that | Now identifying <math>A = W^2</math> and <math>B = j^{\mathsf T}j</math>. We can see that | ||
:<math> | :<math> | ||
G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2} | G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2} | ||
</math> | </math> | ||
Now let <math>l = \begin{bmatrix} | Now let <math> | ||
\frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\ | l = \begin{bmatrix} | ||
\end{bmatrix} </math>, then | \frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\ | ||
\end{bmatrix} | |||
</math>, then | |||
:<math> l = W^{-2}j </math> | |||
:<math> | :<math> | ||
G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l | |||
G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l | |||
</math> | </math> | ||
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:<math> | :<math> | ||
j^{\mathsf T}j \circ W^{-2} = I_n | j^{\mathsf T}j \circ W^{-2} = I_n | ||
g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n | </math> | ||
:<math> | |||
g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n | |||
</math> | </math> | ||
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:<math> | :<math> | ||
G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l | G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l | ||
</math> | </math> | ||
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:<math> | :<math> | ||
\begin{aligned} | \begin{aligned} | ||
G_a(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\ | G_a(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\ | ||
&= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n} | &= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n} | ||
\end{aligned} | \end{aligned} | ||
</math> | </math> | ||
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:<math> | :<math> | ||
\begin{aligned} | \begin{aligned} | ||
G_b(E) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^2) - \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\ | G_b(E) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^2) - \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\ | ||
&= \frac{W^{-2}}{n} (1+E^2) - \frac{l^{\mathsf T}l}{n^2} | &= \frac{W^{-2}}{n} (1+E^2) - \frac{l^{\mathsf T}l}{n^2} | ||
\end{aligned} | \end{aligned} | ||
</math> | </math> | ||
Since the metric is equivalent up to scaling, we multiply by <math>\frac{n}{1+E^ | Since the metric is equivalent up to scaling, we multiply by <math>\frac{n}{1+E^2}</math> to obtain: | ||
:<math> | :<math> | ||
G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l | G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l | ||
</math> | </math> | ||
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:<math> | :<math> | ||
X_k = \begin{bmatrix} | X_k = | ||
W \\ | \begin{bmatrix} | ||
\hline | W \\ | ||
k\cdot j | \hline | ||
\end{bmatrix} | k\cdot j | ||
G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j | \end{bmatrix} | ||
</math> | |||
:<math> | |||
G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j | |||
</math> | </math> | ||
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:<math> | :<math> | ||
G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l | G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l | ||
</math> | </math> | ||
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:<math> | :<math> | ||
\begin{ | \begin{aligned} | ||
nk^2E^2 = 1\\ | nk^2E^2 &= 1\\ | ||
E = \sqrt{\frac{1}{nk^2}}\\ | E &= \sqrt{\frac{1}{nk^2}}\\ | ||
k = \sqrt{\frac{1}{nE^2}} | k &= \sqrt{\frac{1}{nE^2}} | ||
\end{ | \end{aligned} | ||
</math> | </math> | ||
== References == | == References == | ||