Matrix Derivative: Why is GRADIENT f(X) = C?

[strokebow]

Hi,

A function: f(X) = C^T*X

Where, ^T is Transpose


Then my book tells me that GRADIENT f(X) = C


Why? Why is it not GRADIENT f(X) = C^T

Where, ^T is Transpose and GRADIENT is labla (opposite to delta)

Please help!

Thanks

 

[HallsofIvy]

So
C= \begin{bmatrix}a \\ b \\ c\end{bmatrix}
C^T= \begin{bmatrix}a & b & c\end{bmatrix}
and
C^T X= \begin{bmatrix} a & b & c\end{bmatrix}\begin{bmatrix}x \\ y \\ c\end{bmatrix}= ax+ ay+ az

So
\nabla C^T X= \nabla (ax+ ay+ az)= \begin{bmatrix}a \\ b \\ c}\end{bmatrix}= C[/itex]

 

[Hurkyl]

Why? Why is it not GRADIENT f(X) = C^T

Check the definitions; that should make it obvious.

(I should point out that when paying attention to row vectors vs. column vectors, there are two different for defining the gradient, one being the transpose of the other)