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typo and formatting
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edit approved Dec 20, 2017 at 10:10
Dmitri Zaitsev
edit approved Dec 20, 2017 at 10:10
Dmitri Zaitsev
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Let $F = \{\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123}\}$. $$F = \big\{ \Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123} \big\}.$$

The claim in the linked talk is not that $F$ is a linear basis for $\mathbb{C}[Mat_3] = \mathbb{C}[x_{ij}]$, but that each $x_{ij}$ can be expressed as a rational function in the $\Delta^J_I \in F$. The claim is actually a bit more, it is that each minor of the matrix $M = (x_{ij})$ is a subtraction-free rational function in the elements of $F$. This means if each minor in $F$ is positive, then the matrix is totally positive.

Here is how to recover $x_{11}$. Notice we are given $x_{22}, x_{23}, x_{32}, x_{13}$, and $x_{3,1}$$x_{31}$. We can then recover $$x_{33} = \frac{\Delta^{23}_{23} + x_{23}x_{32}}{x_{22}}$$ since we when also given $\Delta^{23}_{23}$. We can then similarly recover $x_{12}$ and $x_{21}$ since we have the minors $\Delta^{23}_{12}$ and $\Delta^{12}_{23}$ along with all but one relevant coordinate in each case. We can then recover $$\Delta_{12}^{12} = \frac{x_{22}\Delta_{123}^{123} + \Delta^{12}_{23}\Delta^{23}_{12}}{\Delta^{23}_{23}}$$ using the Lewis Carroll identity. Finally we get $x_{11}$ using $\Delta_{12}^{12}$ along with $x_{12}, x_{21}$, and $x_{22}$.

Let $F = \{\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123}\}$. The claim in the linked talk is not that $F$ is a linear basis for $\mathbb{C}[Mat_3] = \mathbb{C}[x_{ij}]$, but that each $x_{ij}$ can be expressed as a rational function in the $\Delta^J_I \in F$. The claim is actually a bit more, it is that each minor of the matrix $M = (x_{ij})$ is a subtraction-free rational function in the elements of $F$. This means if each minor in $F$ is positive, then the matrix is totally positive.

Here is how to recover $x_{11}$. Notice we are given $x_{22}, x_{23}, x_{32}, x_{13}$, and $x_{3,1}$. We can then recover $$x_{33} = \frac{\Delta^{23}_{23} + x_{23}x_{32}}{x_{22}}$$ since we when also given $\Delta^{23}_{23}$. We can then similarly recover $x_{12}$ and $x_{21}$ since we have the minors $\Delta^{23}_{12}$ and $\Delta^{12}_{23}$ along with all but one relevant coordinate in each case. We can then recover $$\Delta_{12}^{12} = \frac{x_{22}\Delta_{123}^{123} + \Delta^{12}_{23}\Delta^{23}_{12}}{\Delta^{23}_{23}}$$ using the Lewis Carroll identity. Finally we get $x_{11}$ using $\Delta_{12}^{12}$ along with $x_{12}, x_{21}$, and $x_{22}$.

Let $$F = \big\{ \Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123} \big\}.$$

The claim in the linked talk is not that $F$ is a linear basis for $\mathbb{C}[Mat_3] = \mathbb{C}[x_{ij}]$, but that each $x_{ij}$ can be expressed as a rational function in the $\Delta^J_I \in F$. The claim is actually a bit more, it is that each minor of the matrix $M = (x_{ij})$ is a subtraction-free rational function in the elements of $F$. This means if each minor in $F$ is positive, then the matrix is totally positive.

Here is how to recover $x_{11}$. Notice we are given $x_{22}, x_{23}, x_{32}, x_{13}$, and $x_{31}$. We can then recover $$x_{33} = \frac{\Delta^{23}_{23} + x_{23}x_{32}}{x_{22}}$$ since we when also given $\Delta^{23}_{23}$. We can then similarly recover $x_{12}$ and $x_{21}$ since we have the minors $\Delta^{23}_{12}$ and $\Delta^{12}_{23}$ along with all but one relevant coordinate in each case. We can then recover $$\Delta_{12}^{12} = \frac{x_{22}\Delta_{123}^{123} + \Delta^{12}_{23}\Delta^{23}_{12}}{\Delta^{23}_{23}}$$ using the Lewis Carroll identity. Finally we get $x_{11}$ using $\Delta_{12}^{12}$ along with $x_{12}, x_{21}$, and $x_{22}$.

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John Machacek
John Machacek
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Let $F = \{\Delta_2^2, \Delta_{23}^{23}, \Delta_{3}^{2}, \Delta_{2}^{3}, \Delta_{1}^{3}, \Delta_{3}^{1}, \Delta_{12}^{23}, \Delta_{23}^{12}, \Delta_{123}^{123}\}$. The claim in the linked talk is not that $F$ is a linear basis for $\mathbb{C}[Mat_3] = \mathbb{C}[x_{ij}]$, but that each $x_{ij}$ can be expressed as a rational function in the $\Delta^J_I \in F$. The claim is actually a bit more, it is that each minor of the matrix $M = (x_{ij})$ is a subtraction-free rational function in the elements of $F$. This means if each minor in $F$ is positive, then the matrix is totally positive.

Here is how to recover $x_{11}$. Notice we are given $x_{22}, x_{23}, x_{32}, x_{13}$, and $x_{3,1}$. We can then recover $$x_{33} = \frac{\Delta^{23}_{23} + x_{23}x_{32}}{x_{22}}$$ since we when also given $\Delta^{23}_{23}$. We can then similarly recover $x_{12}$ and $x_{21}$ since we have the minors $\Delta^{23}_{12}$ and $\Delta^{12}_{23}$ along with all but one relevant coordinate in each case. We can then recover $$\Delta_{12}^{12} = \frac{x_{22}\Delta_{123}^{123} + \Delta^{12}_{23}\Delta^{23}_{12}}{\Delta^{23}_{23}}$$ using the Lewis Carroll identity. Finally we get $x_{11}$ using $\Delta_{12}^{12}$ along with $x_{12}, x_{21}$, and $x_{22}$.