# Dimension of gradient operator

Suppose

\begin{align} y&=Ax\\ z&=f(y) \end{align}

Then, is it true that: \begin{align} \nabla_xz&=\nabla_yf(y)\nabla_xy \\ &=\nabla_yf(y)A \end{align}

Dimensions:

• $A: m \times n$
• $x: n \times 1$
• $y: m \times 1$
• $z: 1 \times 1$
• $\nabla_yf(y): 1 \times m$?
• $\nabla_xz: 1 \times n$?
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## 1 Answer

\begin{align} \nabla_xz&=(\nabla_xy)^\text{T}\nabla_yf(y) \\ &=A^\text{T}\nabla_yf(y) \end{align}

Should clarify things, and get $\nabla_xz$ and $\nabla_yf(y)$ into column vectors. The dimensions you've listed are right, with the exception of the transposition of the last two vectors.

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