Derivative with respect to a vector

  • #1

Main Question or Discussion Point

Hi, I have a question about doing derivative with respect to a vector, can someone help please.

Problem:
Suppose A is a (nxn) dimensional symmetric matrix, [tex]\vec{x}[/tex] is a (nx1) column vector.
We know that

[tex]\frac{d A\vec{x}}{d \vec{x}}=A[/tex]

and

[tex]\frac{d \vec{x}^TA\vec{x}}{d \vec{x}}=2A\vec{x}[/tex] ( A is symmetric)

question:

[tex]\frac{d \vec{x}^TA}{d \vec{x}}=?[/tex]

many thanks in advance!
 
Last edited:

Answers and Replies

  • #2
dextercioby
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Do you know how to work with tensor formalism? Using subscripts and superscripts.

Daniel.
 
  • #3
this should be A too. When A is not a function of x then take it out of the derivative and realize that dxi/dxj is simply (d)ij where (d) is the isentropic replacement tensor.
(Sorry I didn't feel like latex today)
 
  • #4
robphy
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As dextercioby suggests, [tex]\frac{d \vec{x}^TA}{d \vec{x}}[/tex] can be written as (assuming that A is independent of x)
[tex]\frac{\partial (x_a A^a{}_b)}{\partial x^c}=\frac{\partial (x_a )}{\partial x^c}A^a{}_b+x_a \frac{\partial (A^a{}_b)}{\partial x^c}
=\delta_{ac}A^a{}_b+0
=A_{cb}
[/tex]
which is not exactly the object [tex]A^a{}_b[/tex] that we started with.

But, now, what is the interpretation of a "derivative with respect to a vector [or a tensor]"?
 

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