Mixing
need help with matrix calculus
$begingroup$
I am trying to find $frac{partial (x'Ax)}{partial x}$ where x is a vector (2 x 1 vector) and A is a matrix (say 2x2 dimensions).
When I looked up in http://www.matrixcalculus.org/ I found the answer to be $(A.x)' + x'.A$ where $'$ stands for transpose.
I tried to solve this using $frac{partial (CB)}{partial x} = frac{partial{C}}{partial x}B + Cfrac{partial B}{partial x}$ where $C = x'A$ and $B = x$. With this in perspective, I am getting the derivative as $A'x + x'A$.
Clearly with $x$ being a column vector and $A$ being a square matrix, my answer is wrong since individual terms ($A'x$ and $x'A$)have different shapes.
Where am I getting the calculations wrong. Any help ?
matrix-calculus
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
$endgroup$
migrated from stats.stackexchange.com Jan 15 at 12:28
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
add a comment |
$begingroup$
I am trying to find $frac{partial (x'Ax)}{partial x}$ where x is a vector (2 x 1 vector) and A is a matrix (say 2x2 dimensions).
When I looked up in http://www.matrixcalculus.org/ I found the answer to be $(A.x)' + x'.A$ where $'$ stands for transpose.
I tried to solve this using $frac{partial (CB)}{partial x} = frac{partial{C}}{partial x}B + Cfrac{partial B}{partial x}$ where $C = x'A$ and $B = x$. With this in perspective, I am getting the derivative as $A'x + x'A$.
Clearly with $x$ being a column vector and $A$ being a square matrix, my answer is wrong since individual terms ($A'x$ and $x'A$)have different shapes.
Where am I getting the calculations wrong. Any help ?
matrix-calculus
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
$endgroup$
migrated from stats.stackexchange.com Jan 15 at 12:28
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
add a comment |
$begingroup$
I am trying to find $frac{partial (x'Ax)}{partial x}$ where x is a vector (2 x 1 vector) and A is a matrix (say 2x2 dimensions).
When I looked up in http://www.matrixcalculus.org/ I found the answer to be $(A.x)' + x'.A$ where $'$ stands for transpose.
I tried to solve this using $frac{partial (CB)}{partial x} = frac{partial{C}}{partial x}B + Cfrac{partial B}{partial x}$ where $C = x'A$ and $B = x$. With this in perspective, I am getting the derivative as $A'x + x'A$.
Clearly with $x$ being a column vector and $A$ being a square matrix, my answer is wrong since individual terms ($A'x$ and $x'A$)have different shapes.
Where am I getting the calculations wrong. Any help ?
matrix-calculus
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
$endgroup$
I am trying to find $frac{partial (x'Ax)}{partial x}$ where x is a vector (2 x 1 vector) and A is a matrix (say 2x2 dimensions).
When I looked up in http://www.matrixcalculus.org/ I found the answer to be $(A.x)' + x'.A$ where $'$ stands for transpose.
I tried to solve this using $frac{partial (CB)}{partial x} = frac{partial{C}}{partial x}B + Cfrac{partial B}{partial x}$ where $C = x'A$ and $B = x$. With this in perspective, I am getting the derivative as $A'x + x'A$.
Clearly with $x$ being a column vector and $A$ being a square matrix, my answer is wrong since individual terms ($A'x$ and $x'A$)have different shapes.
Where am I getting the calculations wrong. Any help ?
matrix-calculus
matrix-calculus
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
asked Jan 15 at 12:27
Upendra Pratap SinghUpendra Pratap Singh
1134
1134
migrated from stats.stackexchange.com Jan 15 at 12:28
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
migrated from stats.stackexchange.com Jan 15 at 12:28
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
add a comment |
add a comment |
1 Answer
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$begingroup$
The rule
$frac{partial mathbf Cmathbf B}{partial x} = frac{partial{mathbf C}}{partial x}mathbf B + mathbf Cfrac{partial mathbf B}{partial x}$
is for differentiation by a scalar $x$ where $mathbf B$ and $mathbf C$ are matrices. It is not a rule for differentiation by a vector.
With $mathbf C = mathbf x^top mathbf A$ and $mathbf B = mathbf x,$ you have a row vector $mathbf C$ and a column vector $mathbf B,$
so you can apply the rule
$$
frac{partial mathbf u^topmathbf v}{partial mathbf x}
= mathbf u^top frac{partial{mathbf v}}{partial mathbf x}
+ mathbf v^top frac{partial{mathbf u}}{partial mathbf x}
$$
with $mathbf u^top = mathbf C$ and $mathbf v = mathbf B,$ so
begin{align}
frac{partial mathbf Cmathbf B}{partial mathbf x}
&= mathbf C frac{partial{mathbf B}}{partial mathbf x}
+ mathbf B^top frac{partial{mathbf C^top}}{partial mathbf x}\
&= mathbf x^top mathbf A frac{partial{mathbf x}}{partial mathbf x}
+ mathbf x^top frac{partial{mathbf A^topmathbf x}}{partial mathbf x}\
&= mathbf x^top mathbf A + mathbf x^top mathbf A^top\
&= mathbf x^top mathbf A + (mathbf Amathbf x)^top.
end{align}
Done!
For differentiation of matrices by vectors, refer to
Derivative of a Matrix with respect to a vector.
But that seems much more than you need or want.
answered Jan 15 at 14:02
David KDavid K
54.5k343119
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
The rule
$frac{partial mathbf Cmathbf B}{partial x} = frac{partial{mathbf C}}{partial x}mathbf B + mathbf Cfrac{partial mathbf B}{partial x}$
is for differentiation by a scalar $x$ where $mathbf B$ and $mathbf C$ are matrices. It is not a rule for differentiation by a vector.
With $mathbf C = mathbf x^top mathbf A$ and $mathbf B = mathbf x,$ you have a row vector $mathbf C$ and a column vector $mathbf B,$
so you can apply the rule
$$
frac{partial mathbf u^topmathbf v}{partial mathbf x}
= mathbf u^top frac{partial{mathbf v}}{partial mathbf x}
+ mathbf v^top frac{partial{mathbf u}}{partial mathbf x}
$$
with $mathbf u^top = mathbf C$ and $mathbf v = mathbf B,$ so
begin{align}
frac{partial mathbf Cmathbf B}{partial mathbf x}
&= mathbf C frac{partial{mathbf B}}{partial mathbf x}
+ mathbf B^top frac{partial{mathbf C^top}}{partial mathbf x}\
&= mathbf x^top mathbf A frac{partial{mathbf x}}{partial mathbf x}
+ mathbf x^top frac{partial{mathbf A^topmathbf x}}{partial mathbf x}\
&= mathbf x^top mathbf A + mathbf x^top mathbf A^top\
&= mathbf x^top mathbf A + (mathbf Amathbf x)^top.
end{align}
Done!
For differentiation of matrices by vectors, refer to
Derivative of a Matrix with respect to a vector.
But that seems much more than you need or want.
answered Jan 15 at 14:02
David KDavid K
54.5k343119
$endgroup$
add a comment |
$begingroup$
The rule
$frac{partial mathbf Cmathbf B}{partial x} = frac{partial{mathbf C}}{partial x}mathbf B + mathbf Cfrac{partial mathbf B}{partial x}$
is for differentiation by a scalar $x$ where $mathbf B$ and $mathbf C$ are matrices. It is not a rule for differentiation by a vector.
With $mathbf C = mathbf x^top mathbf A$ and $mathbf B = mathbf x,$ you have a row vector $mathbf C$ and a column vector $mathbf B,$
so you can apply the rule
$$
frac{partial mathbf u^topmathbf v}{partial mathbf x}
= mathbf u^top frac{partial{mathbf v}}{partial mathbf x}
+ mathbf v^top frac{partial{mathbf u}}{partial mathbf x}
$$
with $mathbf u^top = mathbf C$ and $mathbf v = mathbf B,$ so
begin{align}
frac{partial mathbf Cmathbf B}{partial mathbf x}
&= mathbf C frac{partial{mathbf B}}{partial mathbf x}
+ mathbf B^top frac{partial{mathbf C^top}}{partial mathbf x}\
&= mathbf x^top mathbf A frac{partial{mathbf x}}{partial mathbf x}
+ mathbf x^top frac{partial{mathbf A^topmathbf x}}{partial mathbf x}\
&= mathbf x^top mathbf A + mathbf x^top mathbf A^top\
&= mathbf x^top mathbf A + (mathbf Amathbf x)^top.
end{align}
Done!
For differentiation of matrices by vectors, refer to
Derivative of a Matrix with respect to a vector.
But that seems much more than you need or want.
answered Jan 15 at 14:02
David KDavid K
54.5k343119
$endgroup$
add a comment |
$begingroup$
The rule
$frac{partial mathbf Cmathbf B}{partial x} = frac{partial{mathbf C}}{partial x}mathbf B + mathbf Cfrac{partial mathbf B}{partial x}$
is for differentiation by a scalar $x$ where $mathbf B$ and $mathbf C$ are matrices. It is not a rule for differentiation by a vector.
With $mathbf C = mathbf x^top mathbf A$ and $mathbf B = mathbf x,$ you have a row vector $mathbf C$ and a column vector $mathbf B,$
so you can apply the rule
$$
frac{partial mathbf u^topmathbf v}{partial mathbf x}
= mathbf u^top frac{partial{mathbf v}}{partial mathbf x}
+ mathbf v^top frac{partial{mathbf u}}{partial mathbf x}
$$
with $mathbf u^top = mathbf C$ and $mathbf v = mathbf B,$ so
begin{align}
frac{partial mathbf Cmathbf B}{partial mathbf x}
&= mathbf C frac{partial{mathbf B}}{partial mathbf x}
+ mathbf B^top frac{partial{mathbf C^top}}{partial mathbf x}\
&= mathbf x^top mathbf A frac{partial{mathbf x}}{partial mathbf x}
+ mathbf x^top frac{partial{mathbf A^topmathbf x}}{partial mathbf x}\
&= mathbf x^top mathbf A + mathbf x^top mathbf A^top\
&= mathbf x^top mathbf A + (mathbf Amathbf x)^top.
end{align}
Done!
For differentiation of matrices by vectors, refer to
Derivative of a Matrix with respect to a vector.
But that seems much more than you need or want.
answered Jan 15 at 14:02
David KDavid K
54.5k343119
$endgroup$
The rule
$frac{partial mathbf Cmathbf B}{partial x} = frac{partial{mathbf C}}{partial x}mathbf B + mathbf Cfrac{partial mathbf B}{partial x}$
is for differentiation by a scalar $x$ where $mathbf B$ and $mathbf C$ are matrices. It is not a rule for differentiation by a vector.
With $mathbf C = mathbf x^top mathbf A$ and $mathbf B = mathbf x,$ you have a row vector $mathbf C$ and a column vector $mathbf B,$
so you can apply the rule
$$
frac{partial mathbf u^topmathbf v}{partial mathbf x}
= mathbf u^top frac{partial{mathbf v}}{partial mathbf x}
+ mathbf v^top frac{partial{mathbf u}}{partial mathbf x}
$$
with $mathbf u^top = mathbf C$ and $mathbf v = mathbf B,$ so
begin{align}
frac{partial mathbf Cmathbf B}{partial mathbf x}
&= mathbf C frac{partial{mathbf B}}{partial mathbf x}
+ mathbf B^top frac{partial{mathbf C^top}}{partial mathbf x}\
&= mathbf x^top mathbf A frac{partial{mathbf x}}{partial mathbf x}
+ mathbf x^top frac{partial{mathbf A^topmathbf x}}{partial mathbf x}\
&= mathbf x^top mathbf A + mathbf x^top mathbf A^top\
&= mathbf x^top mathbf A + (mathbf Amathbf x)^top.
end{align}
Done!
For differentiation of matrices by vectors, refer to
Derivative of a Matrix with respect to a vector.
But that seems much more than you need or want.
answered Jan 15 at 14:02
David KDavid K
54.5k343119
edited Jan 15 at 14:08
answered Jan 15 at 14:02
David KDavid K
54.5k343119
answered Jan 15 at 14:02
David KDavid K
54.5k343119
answered Jan 15 at 14:02
David KDavid K
54.5k343119
54.5k343119
add a comment |
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