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January 14th, 2018, 10:21 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 186 Thanks: 2  Another Least Square question..
The partial derivatives of $\displaystyle Ax^2$ with respect to $\displaystyle x_1......x_n$ fill the vector $\displaystyle 2A^TAx$. The derivatives of $\displaystyle 2b^TAx$ fill the vector $\displaystyle 2A^Tb$. So the derivatives of $\displaystyle Ax  b^2$ are zero when ______? My Answer: Ax = b Am I correct? 
January 23rd, 2018, 10:15 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
The following utilizes the Einstein summation convention: sum over repeated integers, for example $\displaystyle b^{2}=b_{i}b_{i}$ $\displaystyle Axb^{2} = [Axb]_{i} [Axb]_{i}= [[Ax]_{i}b_{i}][[Ax]_{i}b_{i}]$ $\displaystyle =[Ax]_{i}[Ax]_{i} 2b_{i}[Ax]_{i}+b^{2}$ $\displaystyle [Ax]_{i}[Ax]_{i} = A_{ij}x_{j} A_{ik}x_{k} = Ax^{2}$ $\displaystyle \frac{\partial Ax^{2}}{\partial x_{m}}= A_{ij}\delta_{jm} A_{ik}x_{k} + A_{ij}x_{j} A_{ik}\delta_{jm} = 2A_{im}A_{ik}x_{k}=2[A^{T}Ax]_{m}$ $\displaystyle 2b_{i}[Ax]_{i}= 2b_{i}A_{ij}x_{j}$ $\displaystyle \frac{\partial 2b_{i}[Ax]_{i}}{\partial x_{m}} = 2b_{i}A_{ij}\delta _{jm}= 2b_{i}A_{im} = 2A^{T}_{mi}b_{i}$ $\displaystyle \frac{\partial Axb^{2}}{\partial x_{m}}=2[A^{T}AxA^{T}b]_{m}=0$, all m, $\displaystyle \rightarrow$ $\displaystyle A^{T}AxA^{T}b=0$ which is the equation for least squares approximation of Ax=b = 0 when it doesn't have a solution, ie, when you minimize the distance between Ax and b. The utility of the Einstein summation convention never ceases to amaze me. I first tried the following but got stuck: $\displaystyle Axb^{2}=[Axb]^{T}[Axb]= [x^{T}A^{T}b^{T}][Axb]$ $\displaystyle \frac{\partial( x^{T}A^{T}Ax)}{\partial x_{i}}$ = ????? For math equations I use: Online LaTeX Equation Editor  create, integrate and download The following, though a little slow and wordy, is very helpful: https://www.khanacademy.org/math/linearalgebra 
January 24th, 2018, 11:44 AM  #3  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100  Quote:
$\displaystyle Ax^{2} = x^{T}[A^{T}A]x = \begin{vmatrix} x_{1} &x_{2} & . & . & x_{n}\end{vmatrix}[A^{T}A]\begin{vmatrix}x_{1}\\ x_{2}\\ .\\ .\\ x_{n}\end{vmatrix}\\ $ $\displaystyle \frac{\partial (Ax^{2})}{\partial x_{m}}=\begin{vmatrix} 0 &. & 1 & . & 0\end{vmatrix}[A^{T}A]\begin{vmatrix}x_{1}\\ x_{2}\\ .\\ .\\ x_{n}\end{vmatrix} +\begin{vmatrix} x_{1} &x_{2} & . & . & x_{n}\end{vmatrix}[A^{T}A]\begin{vmatrix}0\\ .\\ 1\\ .\\ 0\end{vmatrix}\\$, 1 in the m position $\displaystyle =[A^{T}A]_{row m}x + x^{T}[A^{T}A]_{col m}=2[[A^{T}A]x]_{m}\\$ Following through with other term gives, as before. $\displaystyle \frac{\partial Axb^{2}}{\partial x_{m}}=2[A^{T}AxA^{T}b]_{m}=0$, all m, $\displaystyle \rightarrow$ $\displaystyle A^{T}AxA^{T}b=0$ Another good reference on the subject is the great tutorial https://ocw.mit.edu/courses/mathemat...ideolectures/ See Lectures 15 and 16  
February 3rd, 2018, 10:45 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 
Using a different notation: If $\displaystyle x^{T}=[x_{1}\ x_{2}......x_{n}],\ \ \frac{\partial x^{T}}{\partial x_{i}}= [0\ \ 0..1..0]=e_{i}^{T}$ For a vector $\displaystyle x,\ \ x^{2}=x^{T}x=xx^{T}$ $\displaystyle Axb^{2}=[Axb]^{T}[Axb]= [x^{T}A^{T}b^{T}][Axb]$ $\displaystyle =x^{T}A^{T}Axx^{T}A^{T}bb^{T}Ax b^{2}=x^{T}A^{T}Ax2x^{T}A^{T}bb^{2}$ $\displaystyle \frac{\partial() }{\partial x_{_{i}}}=e_{i}^{T}A^{T}Ax+x^{T}A^{T}Ae_{i}2e_{i}^{T}A^{T}b=2e_{i}^{T}[A^{T}AxA^{T}b]=0,$ all i $\displaystyle A^{T}AxA^{T}b=0$ (In regression, you are dealing with a situation where A doesn't have an inverse.) 

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