Given Simultaneous linear equations of the form, $$a_{11}x_1 + a_{12}x_2+a_{13}x_3+\cdots a_{1n}x_n = b_1$$ $$a_{21}x_1 + a_{22}x_2+a_{23}x_3+\cdots a_{2n}x_n=b_2$$ $$a_{31}x_1 + a_{32}x_2+a_{33}x_3+\cdots a_{3n}x_n=b_3$$ $$\vdots$$ $$a_{n1}x_1 + a_{n2}x_2+a_{n3}x_3+\cdots a_{nn}x_n=b_n$$

Can the Newton Raphson's Method use to solve this system of linear equations? Please mention the reasons and possible arguments.

  • $\begingroup$ In the case of a linear function, what is the difference? $\endgroup$ – Matthew Leingang Sep 10 '19 at 11:56
  • $\begingroup$ @MatthewLeingang In the Case of Linear functions, the inverse of Jacobian is a constant matrix. $\endgroup$ – Akash Tadwai Sep 10 '19 at 12:00
  • $\begingroup$ Are you sure? What's the Jacobian of a linear function? $\endgroup$ – Matthew Leingang Sep 10 '19 at 12:02
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    $\begingroup$ Applying the Newton-Raphson method involves calculating the inverse of a function. In this case, it is equivalent to inverting the coefficient matrix. So basically it's the same thing as just solving the system the normal way. $\endgroup$ – Matti P. Sep 10 '19 at 12:04

That just leads you back to solving the same system in the (only) iteration. No point in that.

There are iterative methods to approximate the solution to (large, sparse) linear systems, like relaxation.

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  • $\begingroup$ I know that there are relaxation methods, but I'm just asking that is the process is same as relaxation methods when we are solving a system of linear equations? $x^{k+1}=x^k-[{A^{(k)}}]^{-1} F[x^{(k)}]$, As inverse is a constant matrix does it mean the same? $\endgroup$ – Akash Tadwai Sep 10 '19 at 12:13
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    $\begingroup$ @AkashTadwai Yes. If you want to find a root of $f(x)=Ax-b$, the Jacobian is $J(x)=A$ so the Newton step converges in one step for any initial guess of $x$. $\endgroup$ – Algebraic Pavel Sep 10 '19 at 13:02
  • $\begingroup$ Take a look at quasi-Newton methods for ways to use Newton-like iterations for large equation systems with cheaper updates. $\endgroup$ – vonbrand Sep 10 '19 at 13:23

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