The Derivative of a Vector












0














It is stated in this video that some books attempt to define the derivative of a vector, with the added caveat that tensor notation is a much better approach. Presumably, this is a different concept than the derivative of a vector-valued function, as the latter is uncontroversial and ubiquitous. What is this unpalatable process used by such books to take the derivative of a vector, and a vector transposed?






calculus linear-algebra derivatives vectors transpose







share|cite|improve this question






















  • I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
    – Cuspy Code
    Dec 6 at 17:44












  • The best framework for this is geometric algebra or Clifford Algebra.
    – John L Winters
    Dec 6 at 19:02
















0














It is stated in this video that some books attempt to define the derivative of a vector, with the added caveat that tensor notation is a much better approach. Presumably, this is a different concept than the derivative of a vector-valued function, as the latter is uncontroversial and ubiquitous. What is this unpalatable process used by such books to take the derivative of a vector, and a vector transposed?






calculus linear-algebra derivatives vectors transpose







share|cite|improve this question






















  • I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
    – Cuspy Code
    Dec 6 at 17:44












  • The best framework for this is geometric algebra or Clifford Algebra.
    – John L Winters
    Dec 6 at 19:02














0












0








0







It is stated in this video that some books attempt to define the derivative of a vector, with the added caveat that tensor notation is a much better approach. Presumably, this is a different concept than the derivative of a vector-valued function, as the latter is uncontroversial and ubiquitous. What is this unpalatable process used by such books to take the derivative of a vector, and a vector transposed?






calculus linear-algebra derivatives vectors transpose







share|cite|improve this question













It is stated in this video that some books attempt to define the derivative of a vector, with the added caveat that tensor notation is a much better approach. Presumably, this is a different concept than the derivative of a vector-valued function, as the latter is uncontroversial and ubiquitous. What is this unpalatable process used by such books to take the derivative of a vector, and a vector transposed?






calculus linear-algebra derivatives vectors transpose




calculus linear-algebra derivatives vectors transpose






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asked Dec 6 at 17:29









user10478

388111




388111












  • I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
    – Cuspy Code
    Dec 6 at 17:44












  • The best framework for this is geometric algebra or Clifford Algebra.
    – John L Winters
    Dec 6 at 19:02


















  • I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
    – Cuspy Code
    Dec 6 at 17:44












  • The best framework for this is geometric algebra or Clifford Algebra.
    – John L Winters
    Dec 6 at 19:02
















I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
– Cuspy Code
Dec 6 at 17:44






I didn't click on the youtube link, but as a general comment derivatives are always based on changes in functions, which are then taken to the limit as some parameter approaches zero. Notation doesn't change this, it's more a matter of convenience.
– Cuspy Code
Dec 6 at 17:44














The best framework for this is geometric algebra or Clifford Algebra.
– John L Winters
Dec 6 at 19:02




The best framework for this is geometric algebra or Clifford Algebra.
– John L Winters
Dec 6 at 19:02










1 Answer
1






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oldest

votes


















2














The linked video appears to be a discussion about minimizing
$$eqalign{ f(x) &= tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.

First find the differential and gradient of the function.
$$eqalign{
f &= tfrac{1}{2}x^TAx - b^Tx cr
df &= tfrac{1}{2}(dx^TAx+x^TA,dx) - b^Tdx cr
&= (x^TA -b^T),dx cr
&= (Ax-b)^T,dx cr
frac{partial f}{partial x} &= Ax - b cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.



The same calculation in index/tensor notation looks like
$$eqalign{
f &= tfrac{1}{2}x_iA_{ij}x_j - b_ix_i cr
frac{partial f}{partial x_k}
&= tfrac{1}{2}(delta_{ik}A_{ij}x_j+x_iA_{ij}delta_{jk}) - b_idelta_{ik} cr
&= tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k cr
&= tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k cr
&= A_{kj}x_j - b_k cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.






share|cite|improve this answer





















  • Why does $dx^TAx = x^TAdx$?
    – user10478
    Dec 8 at 17:42










  • Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
    – greg
    Dec 8 at 18:48













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














The linked video appears to be a discussion about minimizing
$$eqalign{ f(x) &= tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.

First find the differential and gradient of the function.
$$eqalign{
f &= tfrac{1}{2}x^TAx - b^Tx cr
df &= tfrac{1}{2}(dx^TAx+x^TA,dx) - b^Tdx cr
&= (x^TA -b^T),dx cr
&= (Ax-b)^T,dx cr
frac{partial f}{partial x} &= Ax - b cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.



The same calculation in index/tensor notation looks like
$$eqalign{
f &= tfrac{1}{2}x_iA_{ij}x_j - b_ix_i cr
frac{partial f}{partial x_k}
&= tfrac{1}{2}(delta_{ik}A_{ij}x_j+x_iA_{ij}delta_{jk}) - b_idelta_{ik} cr
&= tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k cr
&= tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k cr
&= A_{kj}x_j - b_k cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.






share|cite|improve this answer





















  • Why does $dx^TAx = x^TAdx$?
    – user10478
    Dec 8 at 17:42










  • Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
    – greg
    Dec 8 at 18:48


















2














The linked video appears to be a discussion about minimizing
$$eqalign{ f(x) &= tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.

First find the differential and gradient of the function.
$$eqalign{
f &= tfrac{1}{2}x^TAx - b^Tx cr
df &= tfrac{1}{2}(dx^TAx+x^TA,dx) - b^Tdx cr
&= (x^TA -b^T),dx cr
&= (Ax-b)^T,dx cr
frac{partial f}{partial x} &= Ax - b cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.



The same calculation in index/tensor notation looks like
$$eqalign{
f &= tfrac{1}{2}x_iA_{ij}x_j - b_ix_i cr
frac{partial f}{partial x_k}
&= tfrac{1}{2}(delta_{ik}A_{ij}x_j+x_iA_{ij}delta_{jk}) - b_idelta_{ik} cr
&= tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k cr
&= tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k cr
&= A_{kj}x_j - b_k cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.






share|cite|improve this answer





















  • Why does $dx^TAx = x^TAdx$?
    – user10478
    Dec 8 at 17:42










  • Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
    – greg
    Dec 8 at 18:48
















2












2








2






The linked video appears to be a discussion about minimizing
$$eqalign{ f(x) &= tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.

First find the differential and gradient of the function.
$$eqalign{
f &= tfrac{1}{2}x^TAx - b^Tx cr
df &= tfrac{1}{2}(dx^TAx+x^TA,dx) - b^Tdx cr
&= (x^TA -b^T),dx cr
&= (Ax-b)^T,dx cr
frac{partial f}{partial x} &= Ax - b cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.



The same calculation in index/tensor notation looks like
$$eqalign{
f &= tfrac{1}{2}x_iA_{ij}x_j - b_ix_i cr
frac{partial f}{partial x_k}
&= tfrac{1}{2}(delta_{ik}A_{ij}x_j+x_iA_{ij}delta_{jk}) - b_idelta_{ik} cr
&= tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k cr
&= tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k cr
&= A_{kj}x_j - b_k cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.






share|cite|improve this answer












The linked video appears to be a discussion about minimizing
$$eqalign{ f(x) &= tfrac{1}{2}x^TAx - x^Tb}$$
Here's how one might solve it using matrix notation.

First find the differential and gradient of the function.
$$eqalign{
f &= tfrac{1}{2}x^TAx - b^Tx cr
df &= tfrac{1}{2}(dx^TAx+x^TA,dx) - b^Tdx cr
&= (x^TA -b^T),dx cr
&= (Ax-b)^T,dx cr
frac{partial f}{partial x} &= Ax - b cr
}$$
Then set the gradient to zero and solve
$$(Ax-b)=0 implies x=A^{-1}b$$
So it's not really as bad as the lecturer is making it out to be.



The same calculation in index/tensor notation looks like
$$eqalign{
f &= tfrac{1}{2}x_iA_{ij}x_j - b_ix_i cr
frac{partial f}{partial x_k}
&= tfrac{1}{2}(delta_{ik}A_{ij}x_j+x_iA_{ij}delta_{jk}) - b_idelta_{ik} cr
&= tfrac{1}{2}(A_{kj}x_j+x_iA_{ik}) - b_k cr
&= tfrac{1}{2}(A_{kj}x_j+A_{ki}^Tx_i) - b_k cr
&= A_{kj}x_j - b_k cr
}$$
For dealing with scalars, vectors, and matrices (as in this problem) the two notations are roughly equivalent. The real power of index/tensor notation is in dealing with higher-order tensors.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 at 6:46









greg

7,5001721




7,5001721












  • Why does $dx^TAx = x^TAdx$?
    – user10478
    Dec 8 at 17:42










  • Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
    – greg
    Dec 8 at 18:48




















  • Why does $dx^TAx = x^TAdx$?
    – user10478
    Dec 8 at 17:42










  • Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
    – greg
    Dec 8 at 18:48


















Why does $dx^TAx = x^TAdx$?
– user10478
Dec 8 at 17:42




Why does $dx^TAx = x^TAdx$?
– user10478
Dec 8 at 17:42












Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
– greg
Dec 8 at 18:48






Because of two things: 1) $A=A^T$ and 2) for real vectors $x^Ty=y^Tx$
– greg
Dec 8 at 18:48




















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