The derivative of a matrix element with respect to the matrix:

$\displaystyle \frac{d\ x_{ij}}{dX}$

Is there any equation or formula for it?

I could not find it on the Web. If you know any book treating it, I would appreciate the title.

Thanks!

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- Feb 26th 2010, 01:03 PMpaolopiaceDerivative of matrix element
The derivative of a matrix element with respect to the matrix:

$\displaystyle \frac{d\ x_{ij}}{dX}$

Is there any equation or formula for it?

I could not find it on the Web. If you know any book treating it, I would appreciate the title.

Thanks! - Feb 27th 2010, 06:22 AMHallsofIvy
Derivatives only make sense for

**functions**so this matrix and matrix element must have one or more variables. Let's say that each element, and so the matrix, if a function of t. Then $\displaystyle \frac{dx_{ij}}{dt}$ is the derivative of the function $\displaystyle x_{ij}(t)$ and $\displaystyle \frac{dX}{dt}$ is the matrix having those derivatives as elements.

By the chain rule, $\displaystyle \frac{dx_{ij}}{dX}= \frac{dx_{ij}}{dt}\frac{dt}{dX}= \frac{dx_{ij}}{dt}\left(\frac{dX}{dt}\right)^{-1}$.

If that inverse does not exist, the derivative does not exist. - Feb 27th 2010, 08:04 AMpaolopiace
HallsofIvy,

There is no function. In $\displaystyle \frac{d x_{ij}}{d X}$ the matrix X is the variable in a quadratic form. To me, it seems like doing $\displaystyle \frac{d}{d x} x = 1$ in one dimension.

Anyway, I really reach the point where I have $\displaystyle \frac{d x_{ij}}{d X}$.

Does $\displaystyle \frac{d x_{ij}}{d X}$ = 1 make sense? Should I post the whole equation?

Thanks and Regards. - Feb 27th 2010, 08:39 AMOpalg
I agree with HallsofIvy. It is not orthodox mathematics to differentiate with respect to a matrix. Nevertheless, some people have tried to formulate such a concept, and you may find this Wikipedia page informative.

According to that page, if f(X) is a scalar-valued function of an n×m matrix X then the derivative $\displaystyle \frac{df}{dX}$ is defined to be the m×n matrix whose (i,j)-element is $\displaystyle \frac{\partial f}{\partial x_{ji}}$. In particular, if $\displaystyle f(X) = x_{ij}$ then $\displaystyle \frac{dx_{ij}}{dX}$ would be a matrix with a 1 in the (j,i)-position and zeros elsewhere.

But notice that much of the material on that Wikipedia page is disputed. I have no idea how reliable or useful this whole concept is. - Feb 27th 2010, 09:05 AMpaolopiace
Oplag, Thanks.

I see it's better if I post a shorter version of the whole function.

I need to obtain the analytic formula of following derivative:

$\displaystyle \frac{d}{d\Sigma}\left[ b^T \Sigma^{-1} b \right]$

where the dxd matrix Sigma is positive definite and decomposed as $\displaystyle \Sigma = AA^T$

$\displaystyle b = \frac{1}{2}\Sigma_{ii} $ is the n-dimensional vector composed of the diagonal of the matrix Sigma.

Although not much relevant, $\displaystyle \Sigma_{ii}= a_i^T a_i $ where ai is the i-th row of A.

Thanks for any help.

P.S. Actually, like $\displaystyle \frac{d}{dx}x = 1$ it could be that $\displaystyle \frac{d}{dX}X = I$. Thus, $\displaystyle \frac{d}{dX}x_{ij} = 1 $ when i=j. Zero when i<>j.