$f(X^1,…,X^n)=det[X^1…X^n]$ is differentiable
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Let $f:mathbb{R}^ntimesdots timesmathbb{R}^nrightarrowmathbb{R}$ defined by $$ f(X^1,...,X^n)=det[X^1...X^n] \ $$ is a differentiable function and $Df_H(A)=sum_{i=1}^ndet[A^1dots A^{i-1}H^iA^{i+1}dots A^n]$.
I wanted to start with showing $DF(A)$ is linear, but I couldn't show it because if I define $T(A)=sumdet[A^1dots H^idots A^n]$, I get
$$
T(alpha A)=sum det[alpha A^1dotsalpha H^idotsalpha A^n]=alpha^nsum[A^1dots H^idots A^n]ne alpha T(A) \
$$
After I show this, I couldn't either to prove that the limit
$$
lim_{Hrightarrow 0}frac{f(A+H)-f(A)-T_H(A)}{left | H right |} \
=lim_{Hrightarrow 0}frac{det[A^1+H^1...A^n+H^n]-det[A^1...A^n]-sumleft | H right |cdotdet [A^1...H^i...A^n]}{left | H right |} \
$$
equals to $0$.
calculus matrices differential-equations determinant
edited Nov 13 at 10:31
Bernard
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asked Nov 13 at 10:26
J. Doe
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up vote
1
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Let $f:mathbb{R}^ntimesdots timesmathbb{R}^nrightarrowmathbb{R}$ defined by $$ f(X^1,...,X^n)=det[X^1...X^n] \ $$ is a differentiable function and $Df_H(A)=sum_{i=1}^ndet[A^1dots A^{i-1}H^iA^{i+1}dots A^n]$.
I wanted to start with showing $DF(A)$ is linear, but I couldn't show it because if I define $T(A)=sumdet[A^1dots H^idots A^n]$, I get
$$
T(alpha A)=sum det[alpha A^1dotsalpha H^idotsalpha A^n]=alpha^nsum[A^1dots H^idots A^n]ne alpha T(A) \
$$
After I show this, I couldn't either to prove that the limit
$$
lim_{Hrightarrow 0}frac{f(A+H)-f(A)-T_H(A)}{left | H right |} \
=lim_{Hrightarrow 0}frac{det[A^1+H^1...A^n+H^n]-det[A^1...A^n]-sumleft | H right |cdotdet [A^1...H^i...A^n]}{left | H right |} \
$$
equals to $0$.
calculus matrices differential-equations determinant
edited Nov 13 at 10:31
Bernard
115k637108
asked Nov 13 at 10:26
J. Doe
83
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
2
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f:mathbb{R}^ntimesdots timesmathbb{R}^nrightarrowmathbb{R}$ defined by $$ f(X^1,...,X^n)=det[X^1...X^n] \ $$ is a differentiable function and $Df_H(A)=sum_{i=1}^ndet[A^1dots A^{i-1}H^iA^{i+1}dots A^n]$.
I wanted to start with showing $DF(A)$ is linear, but I couldn't show it because if I define $T(A)=sumdet[A^1dots H^idots A^n]$, I get
$$
T(alpha A)=sum det[alpha A^1dotsalpha H^idotsalpha A^n]=alpha^nsum[A^1dots H^idots A^n]ne alpha T(A) \
$$
After I show this, I couldn't either to prove that the limit
$$
lim_{Hrightarrow 0}frac{f(A+H)-f(A)-T_H(A)}{left | H right |} \
=lim_{Hrightarrow 0}frac{det[A^1+H^1...A^n+H^n]-det[A^1...A^n]-sumleft | H right |cdotdet [A^1...H^i...A^n]}{left | H right |} \
$$
equals to $0$.
calculus matrices differential-equations determinant
edited Nov 13 at 10:31
Bernard
115k637108
asked Nov 13 at 10:26
J. Doe
83
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $f:mathbb{R}^ntimesdots timesmathbb{R}^nrightarrowmathbb{R}$ defined by $$ f(X^1,...,X^n)=det[X^1...X^n] \ $$ is a differentiable function and $Df_H(A)=sum_{i=1}^ndet[A^1dots A^{i-1}H^iA^{i+1}dots A^n]$.
I wanted to start with showing $DF(A)$ is linear, but I couldn't show it because if I define $T(A)=sumdet[A^1dots H^idots A^n]$, I get
$$
T(alpha A)=sum det[alpha A^1dotsalpha H^idotsalpha A^n]=alpha^nsum[A^1dots H^idots A^n]ne alpha T(A) \
$$
After I show this, I couldn't either to prove that the limit
$$
lim_{Hrightarrow 0}frac{f(A+H)-f(A)-T_H(A)}{left | H right |} \
=lim_{Hrightarrow 0}frac{det[A^1+H^1...A^n+H^n]-det[A^1...A^n]-sumleft | H right |cdotdet [A^1...H^i...A^n]}{left | H right |} \
$$
equals to $0$.
calculus matrices differential-equations determinant
calculus matrices differential-equations determinant
edited Nov 13 at 10:31
Bernard
115k637108
asked Nov 13 at 10:26
J. Doe
83
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 13 at 10:31
Bernard
115k637108
asked Nov 13 at 10:26
J. Doe
83
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J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 13 at 10:31
Bernard
115k637108
edited Nov 13 at 10:31
Bernard
115k637108
edited Nov 13 at 10:31
Bernard
115k637108
115k637108
asked Nov 13 at 10:26
J. Doe
83
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 10:26
J. Doe
83
asked Nov 13 at 10:26
J. Doe
83
83
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
2
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03
add a comment |
Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
2
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03
Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
2
2
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03
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J. Doe is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to Maths SX! Does $;X^1…X^n;$ denote the product componentwise?
– Bernard
Nov 13 at 10:35
@Bernard Thank you! And yes
– J. Doe
Nov 13 at 10:41
Maybe this helps: math.stackexchange.com/questions/2989656/…
– Christian Blatter
Nov 13 at 10:56
2
Please fix and explain your notations. For my taste, i would write $f$ for the original multilinear map, $Df = f'$ for the differential, $Df(A)=f'(A)$ for the differential, computed as a linear map from the (tangential space in $A$ for the) space $(Bbb R^n)^n$, and $Df(A)(H)$ for ist value in $H$. Now note that you have to prove only the linearity in $H$, $A$ remains fixed, so do not put a scalar also near it...
– dan_fulea
Nov 13 at 11:03