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Derivative of a multivariate quadratic
Direction for greatest derivativeTaking derivative with respect to a vectorThe derivative of a linear operator with respect to its argumentdifficulty with the derivative of L2 normWhat's the most fundamental derivative of multivariable functions?Derivative of $f(X)=X^T$Derivative of transpose of a matrixderivative for a vectorDerivative $frac{partial}{partial X^H} |Tr(X X^top)|^2$Taking partial derivative of matrix
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 14:29
ProgBProgB
31
asked Mar 10 at 14:29
ProgBProgB
31
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The function is
$$F = frac12 boldsymbol{x}^Tboldsymbol{x}-boldsymbol{b}^Tboldsymbol{x}.$$
The total derivative with respect to $boldsymbol{x}$ is given as
$$dF = frac12 dboldsymbol{x}^Tboldsymbol{x}+frac12 boldsymbol{x}^Tdboldsymbol{x}-boldsymbol{b}^Tdboldsymbol{x}$$
$$implies dF = dboldsymbol{x}^Tleft[boldsymbol{x}-boldsymbol{b} right].$$
A $dF = dboldsymbol{x}^Ttext{grad}F$ comparison yields
$$text{grad}F = boldsymbol{x}-boldsymbol{b}.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function is
$$F = frac12 boldsymbol{x}^Tboldsymbol{x}-boldsymbol{b}^Tboldsymbol{x}.$$
The total derivative with respect to $boldsymbol{x}$ is given as
$$dF = frac12 dboldsymbol{x}^Tboldsymbol{x}+frac12 boldsymbol{x}^Tdboldsymbol{x}-boldsymbol{b}^Tdboldsymbol{x}$$
$$implies dF = dboldsymbol{x}^Tleft[boldsymbol{x}-boldsymbol{b} right].$$
A $dF = dboldsymbol{x}^Ttext{grad}F$ comparison yields
$$text{grad}F = boldsymbol{x}-boldsymbol{b}.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
The function is
$$F = frac12 boldsymbol{x}^Tboldsymbol{x}-boldsymbol{b}^Tboldsymbol{x}.$$
The total derivative with respect to $boldsymbol{x}$ is given as
$$dF = frac12 dboldsymbol{x}^Tboldsymbol{x}+frac12 boldsymbol{x}^Tdboldsymbol{x}-boldsymbol{b}^Tdboldsymbol{x}$$
$$implies dF = dboldsymbol{x}^Tleft[boldsymbol{x}-boldsymbol{b} right].$$
A $dF = dboldsymbol{x}^Ttext{grad}F$ comparison yields
$$text{grad}F = boldsymbol{x}-boldsymbol{b}.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
The function is
$$F = frac12 boldsymbol{x}^Tboldsymbol{x}-boldsymbol{b}^Tboldsymbol{x}.$$
The total derivative with respect to $boldsymbol{x}$ is given as
$$dF = frac12 dboldsymbol{x}^Tboldsymbol{x}+frac12 boldsymbol{x}^Tdboldsymbol{x}-boldsymbol{b}^Tdboldsymbol{x}$$
$$implies dF = dboldsymbol{x}^Tleft[boldsymbol{x}-boldsymbol{b} right].$$
A $dF = dboldsymbol{x}^Ttext{grad}F$ comparison yields
$$text{grad}F = boldsymbol{x}-boldsymbol{b}.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
The function is
$$F = frac12 boldsymbol{x}^Tboldsymbol{x}-boldsymbol{b}^Tboldsymbol{x}.$$
The total derivative with respect to $boldsymbol{x}$ is given as
$$dF = frac12 dboldsymbol{x}^Tboldsymbol{x}+frac12 boldsymbol{x}^Tdboldsymbol{x}-boldsymbol{b}^Tdboldsymbol{x}$$
$$implies dF = dboldsymbol{x}^Tleft[boldsymbol{x}-boldsymbol{b} right].$$
A $dF = dboldsymbol{x}^Ttext{grad}F$ comparison yields
$$text{grad}F = boldsymbol{x}-boldsymbol{b}.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
edited Mar 10 at 14:44
J.G.
29.9k22947
edited Mar 10 at 14:44
J.G.
29.9k22947
edited Mar 10 at 14:44
J.G.
29.9k22947
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
5927
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
$boldsymbol{b}^Tdboldsymbol{x}$ is a scalar. Hence, we can transpose it $dboldsymbol{x}^Tboldsymbol{b}$ to obtain the same scalar. Then factor $dboldsymbol{x}^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
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