# Jacobian of a linear transformation

I am currently solving a problem where I have to use the transformation:

\begin{align*}y_1 &=\bar{x}\\ y_2 &= x_2-\bar{x}\\ y_3 &= x_3-\bar{x}\\ &\vdots \\ y_n &= x_n-\bar{x}\\ \end{align*}

My question is: how do I find the Jacobian of the above transformation.

thank you.

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Is $\bar{x}$ constant, or (as I suspect) in terms of the $x_i$? –  Guess who it is. Aug 11 '11 at 16:11
I guess it's in terms of the $x_i$ –  Derek Aug 11 '11 at 16:14
In that case, expand $\bar x$ and compute the Jacobian like usual. For instance, $\frac{\partial x_j}{\partial x_i}=\delta_{ij}$... –  Guess who it is. Aug 11 '11 at 16:17
what if $\bar{x}$ was constant. –  Derek Aug 11 '11 at 16:25
Then the Jacobian is quite trivial... what is the derivative of a constant? –  Guess who it is. Aug 11 '11 at 16:27

This is essentially what Sasha said:

We interpret $\bar x$ as ${1\over n}(x_1+\ldots + x_n)$. Then the Jacobian $J:=\bigl[{\partial y_i\over\partial x_k}\bigr]_{i,k}$ looks as follows: $$\left[\matrix{{1\over n}&{1\over n}&\ldots&{1\over n}\cr -{1\over n}&1-{1\over n}&\ldots&-{1\over n}\cr \vdots\cr -{1\over n}&-{1\over n}&\ldots&1-{1\over n}\cr}\right]\ .$$ If you add the first row to all the other rows, which does not change the determinant, you are left with the matrix $$\left[\matrix{{1\over n}&{1\over n}&\ldots&{1\over n}\cr 0&1&\ldots&0\cr \vdots\cr 0&0&\ldots&1\cr}\right]\ ,$$ whose determinant is obviously ${1\over n}$.

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Thank you Christian. –  Derek Aug 12 '11 at 2:52

For the purpose of determinant computation one can drop subtraction of $\partial_{x_i} \bar{x}$ for $i>1$. Then you are to compute determinant of an identity matrix with first row replaced with constant row of $\frac{1}{n}$. It's determinant is trivially $\frac{1}{n}$ then.

The reason why $\bar{x}$ can be dropped from $y_i$, $i>2$ is that doing so you effectively would subtract from each row below the first the value of the first row, which does not change the determinant.

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Thanks Sasha... –  Derek Aug 12 '11 at 2:52