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

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SUMMARY

The gradient of the function f(X) = C^T * X, where C is a column vector, is definitively equal to C, not C^T. This is due to the definition of the gradient in relation to vector calculus, where the gradient of a scalar function with respect to a vector yields a column vector. The discussion emphasizes the importance of distinguishing between row vectors and column vectors when calculating gradients, as they yield different results based on their orientation.

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strokebow
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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
 
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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]
 
strokebow said:
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)
 

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