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I know the derivative of $(Ax-b)^T(Ax-b)$ is $A^TAx-A^Tb$, but it seems something wrong with my own derivation. Please help me find the error. Thanks

$$ (Ax-b)^T(Ax-b)=(x^TA^T-b^T)(Ax-b)=x^TA^TAx-x^TA^Tb-b^TAx+b^Tb$$

So, taking the derivative with all four terms respect to $x$, we get

$$ (A^TA+(A^TA)^T)x-A^Tb-b^TA+0 $$

what's wrong there?


Edit: with Siong Thye Goh and Bernard's help I got the error, and would like to provide the right derivation here for my own future reference. $$ \begin{align} (Ax-b)^T(Ax-b)&=(x^TA^T-b^T)(Ax-b)\\ &=x^TA^TAx-x^TA^Tb-b^TAx+b^Tb \\ &=x^T(A^TA)x-x^T(A^Tb)-(b^TA)x+b^Tb \end{align} $$

We need to recall two rules:

$$\frac {\partial(a^Tx)} {\partial x}=\frac {\partial(x^Ta)} {\partial x}=a$$

$$\frac {\partial(x^TAx)} {\partial x}=(A+A^T)x$$

Then, take the derivative respect to $x$ we have

$$ \begin{align} (A^TA+(A^TA)^T)x-2A^Tb&=2A^TAx-2A^Tb\\ &=2A^T(Ax-b) \end{align} $$

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    $\begingroup$ If you mean the derivative with respect to $x$ (as opposed to that with respect to $b$) it would be a good idea to say so. If you're looking for the value of $x$ that minimizes the product, that can be done without finding the derivative. Others please note in case you're not aware of it: this product is typically considered when $A$ is a matrix having many more rows than columns, so that $b$ is not generally in the column space of $A$. $\qquad$ $\endgroup$ Aug 10 '16 at 1:32
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Note the derivative of $Ax-b$ is $A$ and the derivative of $\;{}^{\mathrm t\mkern-1.5mu}(Ax-b)$ is $\;{}^{\mathrm t\mkern-1.5mu}A$ since it is the composition of $Ax-b$ with the linear operator of transposition. Thus the derivative of the given expression is \begin{align*} {}^{\mathrm t\!}A(Ax-b)+{}^{\mathrm t\mkern-2mu}(Ax-b)A&= ({}^{\mathrm t\!}A A)x-{}^{\mathrm t\!}Ab+{}^{\mathrm t\mkern-2mu}x({}^{\mathrm t\!}AA)-. \end{align*} Now observe that, since ${}^{\mathrm t\mkern-3mu}AA$ is symmetric, $\;{}^{\mathrm t\mkern-2mu}x({}^{\mathrm t\!}AA)=({}^{\mathrm t\!}AA)x$. Thus the derivative is $$2{\,}^{\mathrm t\!}AAx-{}^{\mathrm t\!}Ab-{}^{\mathrm t\mkern-1mu}bA.$$

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The expected answer is wrong to begin with, it should be $$2(A^TAx-A^Tb)$$

Also, the sizes of $A^Tb$ and $b^TA$ are different, aren't they?

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  • $\begingroup$ thanks for catching this error. I forget the constant 2 because that does not affect setting derivative to 0. Could you tell me more about the error on $A^Tb$ and $b^TA$ thanks again!! $\endgroup$
    – hxd1011
    Aug 10 '16 at 1:53
  • $\begingroup$ $A^Tb$ is column vector but $b^TA$ is a row vector, so they can't be subtracted. $\endgroup$ Aug 10 '16 at 2:05
  • $\begingroup$ but $x^TA^Tb=b^TAx$, so I believe you can solve the problem. $\endgroup$ Aug 10 '16 at 2:06
  • $\begingroup$ Thanks. I think I get it now! $\endgroup$
    – hxd1011
    Aug 10 '16 at 2:07

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