I'm reading the Pattern Recognition and Machine Learning book by Christopher Bishop and I'm having trouble understanding one thing.

Basically, he says:

Given a transformation from $\boldsymbol{x}$ to $\boldsymbol{y}$ defined as $\boldsymbol{y} = (y_{1}, ..., y_{D})$ where $y_{i}=\boldsymbol{u}_{i}^T(\boldsymbol{x} - \boldsymbol{\mu})$ and $\{\boldsymbol{u}_i\}$ is an orthonormal basis, we have a Jacobian matrix $\boldsymbol{J}$ with elements defined as

$$J_{ij} = \frac{\partial x_i}{\partial y_j} = U_{ji} \qquad (1)$$

where $U_{ji}$ are the elements of the matrix $\boldsymbol{U}^T$, which is a matrix of the orthonormal basis vectors $\boldsymbol{u}_i$ as columns.

I don't understand why (1) is true. How come the elements of the Jacobian are equal to the elements of $\boldsymbol{U}^T$?


1 Answer 1


Let's use the notation $\mathbf{u}_i = (U_{i1}, \dots, U_{iD})^T$. Then if $U$ has $\mathbf{u}_i$ as columns, then $U^T$ has entries $U_{ij}$.

Writing the expressions for $y_i$ in coordinates, we get

$$ y_i = \mathbf{u}_i^T (\mathbf{x} - \boldsymbol{\mu}) = \sum_j U_{ij}(x_j-\mu_j) = \sum_j U_{ij} x_j - \sum_j U_{ij} \mu_j $$

I assume $\boldsymbol{\mu}$ is a vector of constants, so the second sum above doesn't contribute to any derivatives. The first sum is just a linear combination of $x_j$, so the coefficients in the Jacobian are

$$ \frac{\partial y_i}{\partial x_j} = U_{ij} $$


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