Matrix quadratic form expansion question












2












$begingroup$


I'm trying to do a question and within it, I need to expand a matrix quadratic form:



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x})$



In my working out, I think that the following is correct:



$$
begin{align}
frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) & = frac{1}{2}(vec{y}^{T} - vec{x}^{T}) Sigma (vec{y} - vec{x}) \
& = frac{1}{2} vec{y}^{T}Sigmavec{y} - frac{1}{2} vec{y}^{T}Sigmavec{x} - frac{1}{2} vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}
end{align}
$$



However, in the answers, it says that the answer is



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} - vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}$



so the middle two cross product terms do not have a half multiplied to them. Can anyone explain this? Or are the answers wrong?



Thanks in advance!






matrices matrix-equations







share|cite|improve this question









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  • $begingroup$
    It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
    $endgroup$
    – amd
    Jan 8 at 19:35
















2












$begingroup$


I'm trying to do a question and within it, I need to expand a matrix quadratic form:



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x})$



In my working out, I think that the following is correct:



$$
begin{align}
frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) & = frac{1}{2}(vec{y}^{T} - vec{x}^{T}) Sigma (vec{y} - vec{x}) \
& = frac{1}{2} vec{y}^{T}Sigmavec{y} - frac{1}{2} vec{y}^{T}Sigmavec{x} - frac{1}{2} vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}
end{align}
$$



However, in the answers, it says that the answer is



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} - vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}$



so the middle two cross product terms do not have a half multiplied to them. Can anyone explain this? Or are the answers wrong?



Thanks in advance!






matrices matrix-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
    $endgroup$
    – amd
    Jan 8 at 19:35














2












2








2





$begingroup$


I'm trying to do a question and within it, I need to expand a matrix quadratic form:



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x})$



In my working out, I think that the following is correct:



$$
begin{align}
frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) & = frac{1}{2}(vec{y}^{T} - vec{x}^{T}) Sigma (vec{y} - vec{x}) \
& = frac{1}{2} vec{y}^{T}Sigmavec{y} - frac{1}{2} vec{y}^{T}Sigmavec{x} - frac{1}{2} vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}
end{align}
$$



However, in the answers, it says that the answer is



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} - vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}$



so the middle two cross product terms do not have a half multiplied to them. Can anyone explain this? Or are the answers wrong?



Thanks in advance!






matrices matrix-equations







share|cite|improve this question









$endgroup$




I'm trying to do a question and within it, I need to expand a matrix quadratic form:



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x})$



In my working out, I think that the following is correct:



$$
begin{align}
frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) & = frac{1}{2}(vec{y}^{T} - vec{x}^{T}) Sigma (vec{y} - vec{x}) \
& = frac{1}{2} vec{y}^{T}Sigmavec{y} - frac{1}{2} vec{y}^{T}Sigmavec{x} - frac{1}{2} vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}
end{align}
$$



However, in the answers, it says that the answer is



$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} - vec{x}^{T}Sigmavec{y} + frac{1}{2}vec{x}^{T}Sigmavec{x}$



so the middle two cross product terms do not have a half multiplied to them. Can anyone explain this? Or are the answers wrong?



Thanks in advance!






matrices matrix-equations




matrices matrix-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 8 at 18:02









user-2147482565user-2147482565

133




133












  • $begingroup$
    It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
    $endgroup$
    – amd
    Jan 8 at 19:35


















  • $begingroup$
    It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
    $endgroup$
    – amd
    Jan 8 at 19:35
















$begingroup$
It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
$endgroup$
– amd
Jan 8 at 19:35




$begingroup$
It looks like whoever wrote that answer had in mind combining the cross terms, which one can do when $Sigma$ is symmetric, but wrote something else.
$endgroup$
– amd
Jan 8 at 19:35










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your answer is correct. Also if $Sigma$ is a symmetric matrix you can simplify the expression as $$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} + frac{1}{2}vec{x}^{T}Sigmavec{x}$$since $$vec{x}^{T}Sigmavec{y}=vec{y}^{T}Sigmavec{x}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you! I was just wanting to double check. I was so confused for a while
    $endgroup$
    – user-2147482565
    Jan 8 at 18:20










  • $begingroup$
    Good luck!.....
    $endgroup$
    – Mostafa Ayaz
    Jan 8 at 18:26



















1












$begingroup$

Seems to me like the from your Book (?) is wrong and yours is correct.



A good way to do a sanity check for such results is to look at the one-dimensional case. If you then further set $Sigma = 1$, the expression you are trying to expand is $frac12(y - x)(y-x) = frac12(y^2 - 2 x y + x^2) = frac12 y^2 - frac12 y x - frac12 x y + frac12 x^2$, as you calculated. Note that this is not equal to $frac12y^2 - 2 x y + frac12 x^2$ (for example, set $x = y = 1$), the result you Book (?) implies.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the reply! This is really helpful
    $endgroup$
    – user-2147482565
    Jan 8 at 18:21












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your answer is correct. Also if $Sigma$ is a symmetric matrix you can simplify the expression as $$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} + frac{1}{2}vec{x}^{T}Sigmavec{x}$$since $$vec{x}^{T}Sigmavec{y}=vec{y}^{T}Sigmavec{x}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you! I was just wanting to double check. I was so confused for a while
    $endgroup$
    – user-2147482565
    Jan 8 at 18:20










  • $begingroup$
    Good luck!.....
    $endgroup$
    – Mostafa Ayaz
    Jan 8 at 18:26
















1












$begingroup$

Your answer is correct. Also if $Sigma$ is a symmetric matrix you can simplify the expression as $$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} + frac{1}{2}vec{x}^{T}Sigmavec{x}$$since $$vec{x}^{T}Sigmavec{y}=vec{y}^{T}Sigmavec{x}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you! I was just wanting to double check. I was so confused for a while
    $endgroup$
    – user-2147482565
    Jan 8 at 18:20










  • $begingroup$
    Good luck!.....
    $endgroup$
    – Mostafa Ayaz
    Jan 8 at 18:26














1












1








1





$begingroup$

Your answer is correct. Also if $Sigma$ is a symmetric matrix you can simplify the expression as $$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} + frac{1}{2}vec{x}^{T}Sigmavec{x}$$since $$vec{x}^{T}Sigmavec{y}=vec{y}^{T}Sigmavec{x}$$






share|cite|improve this answer









$endgroup$



Your answer is correct. Also if $Sigma$ is a symmetric matrix you can simplify the expression as $$frac{1}{2}(vec{y} - vec{x})^{T} Sigma (vec{y} - vec{x}) = frac{1}{2} vec{y}^{T}Sigmavec{y} - vec{y}^{T}Sigmavec{x} + frac{1}{2}vec{x}^{T}Sigmavec{x}$$since $$vec{x}^{T}Sigmavec{y}=vec{y}^{T}Sigmavec{x}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 18:12









Mostafa AyazMostafa Ayaz

18.2k31040




18.2k31040












  • $begingroup$
    Okay thank you! I was just wanting to double check. I was so confused for a while
    $endgroup$
    – user-2147482565
    Jan 8 at 18:20










  • $begingroup$
    Good luck!.....
    $endgroup$
    – Mostafa Ayaz
    Jan 8 at 18:26


















  • $begingroup$
    Okay thank you! I was just wanting to double check. I was so confused for a while
    $endgroup$
    – user-2147482565
    Jan 8 at 18:20










  • $begingroup$
    Good luck!.....
    $endgroup$
    – Mostafa Ayaz
    Jan 8 at 18:26
















$begingroup$
Okay thank you! I was just wanting to double check. I was so confused for a while
$endgroup$
– user-2147482565
Jan 8 at 18:20




$begingroup$
Okay thank you! I was just wanting to double check. I was so confused for a while
$endgroup$
– user-2147482565
Jan 8 at 18:20












$begingroup$
Good luck!.....
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:26




$begingroup$
Good luck!.....
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:26











1












$begingroup$

Seems to me like the from your Book (?) is wrong and yours is correct.



A good way to do a sanity check for such results is to look at the one-dimensional case. If you then further set $Sigma = 1$, the expression you are trying to expand is $frac12(y - x)(y-x) = frac12(y^2 - 2 x y + x^2) = frac12 y^2 - frac12 y x - frac12 x y + frac12 x^2$, as you calculated. Note that this is not equal to $frac12y^2 - 2 x y + frac12 x^2$ (for example, set $x = y = 1$), the result you Book (?) implies.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the reply! This is really helpful
    $endgroup$
    – user-2147482565
    Jan 8 at 18:21
















1












$begingroup$

Seems to me like the from your Book (?) is wrong and yours is correct.



A good way to do a sanity check for such results is to look at the one-dimensional case. If you then further set $Sigma = 1$, the expression you are trying to expand is $frac12(y - x)(y-x) = frac12(y^2 - 2 x y + x^2) = frac12 y^2 - frac12 y x - frac12 x y + frac12 x^2$, as you calculated. Note that this is not equal to $frac12y^2 - 2 x y + frac12 x^2$ (for example, set $x = y = 1$), the result you Book (?) implies.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the reply! This is really helpful
    $endgroup$
    – user-2147482565
    Jan 8 at 18:21














1












1








1





$begingroup$

Seems to me like the from your Book (?) is wrong and yours is correct.



A good way to do a sanity check for such results is to look at the one-dimensional case. If you then further set $Sigma = 1$, the expression you are trying to expand is $frac12(y - x)(y-x) = frac12(y^2 - 2 x y + x^2) = frac12 y^2 - frac12 y x - frac12 x y + frac12 x^2$, as you calculated. Note that this is not equal to $frac12y^2 - 2 x y + frac12 x^2$ (for example, set $x = y = 1$), the result you Book (?) implies.






share|cite|improve this answer









$endgroup$



Seems to me like the from your Book (?) is wrong and yours is correct.



A good way to do a sanity check for such results is to look at the one-dimensional case. If you then further set $Sigma = 1$, the expression you are trying to expand is $frac12(y - x)(y-x) = frac12(y^2 - 2 x y + x^2) = frac12 y^2 - frac12 y x - frac12 x y + frac12 x^2$, as you calculated. Note that this is not equal to $frac12y^2 - 2 x y + frac12 x^2$ (for example, set $x = y = 1$), the result you Book (?) implies.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 18:08









0x5390x539

1,445518




1,445518












  • $begingroup$
    Thanks for the reply! This is really helpful
    $endgroup$
    – user-2147482565
    Jan 8 at 18:21


















  • $begingroup$
    Thanks for the reply! This is really helpful
    $endgroup$
    – user-2147482565
    Jan 8 at 18:21
















$begingroup$
Thanks for the reply! This is really helpful
$endgroup$
– user-2147482565
Jan 8 at 18:21




$begingroup$
Thanks for the reply! This is really helpful
$endgroup$
– user-2147482565
Jan 8 at 18:21


















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