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but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answerthis answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

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rych
rych
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but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto dA^+$$H\mapsto [dA^+](H)$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto dA^+$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto [dA^+](H)$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$

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rych
rych
  • 346
  • 1
  • 2
  • 8

but I can't see how to input the original matrix.

I think when people write $\tfrac {\partial f} { \partial X}$ or $\tfrac {df}{dX}$ they really mean to find the Fréchet derivative, denoted by $df$, or $Df$. Your formula becomes (appears also in this answer) $$\mathrm d A^+ = -A^+ \left( \mathrm d A \right) A^+ +A^+ A{^+}^T \left( \mathrm d A^T \right) (1-A A^+) + (1-A^+ A) \left( \mathrm d A^T \right) A{^+}^T A^+$$

Since your differentiation is with respect to $A$ itself as a variable, then no chain rule needed, $dA=id$ and the derivative (at the given element $A$) is a linear map $H\mapsto dA^+$:

$$[\mathrm d A^+](H) = -A^+ \left( H \right) A^+ +A^+ A{^+}^T \left( H^T \right) (1-A A^+) + (1-A^+ A) \left( H^T \right) A{^+}^T A^+$$