# Why does the Newton-Raphson method sometimes not converge?

I'm talking about the Newton-Raphson method for finding square roots of matrices: $$\begin{cases} X_{k+1} &= \frac{1}{2}\big(X_k + X_k^{-1}A\big)\\ X_0 &= A \end{cases}$$ as a "solution" to the equation $$f(X)=X^2-A=O$$.
For normal functions $$f(x)=x^2-a=0$$ it makes sense why sometimes the method does not converge. But I don't know how to intuitively explain or even understand why the matrix-version sometimes diverges.
For example: compared to the Denman-Beavers method, the Newton-Raphson method is pretty bad since it often does not converge.
So how do we explain these divergences?

One possible way to see what you ask:

Newton's method to solve an equation $$f(u)=0$$ is given by a minimization direction procedure to solve the problem $$\min_u\frac{1}{2}\|f(u)\|^2.$$

You can see this as the initial value problem $$\tag{1}\frac{du}{dt}=-[Jf(u)]^{-1}u,\quad u(0)=u_0,$$ because this implies that
$$\frac{d}{dt}\left(\frac{1}{2}\|f(u)\|^2\right)=-\|f(u)\|^2\leq 0.$$

If you use Euler method to solve (1) numerically, with step size equal $$1$$, you find Newton's method $$\left\{\begin{array}{rll}J{ f}({ u}_j) { w}_j&=&- { f}({ u}_j)\\ { u}_{j+1}&=&{ u}_j+{ w}_j\end{array}\right.$$

Now, if you let $$f(X)=X^2-A,$$ where $$A$$ and $$X$$ are matrix, you find that $$J{ f}({ X}_j) { W}_j=X_jW_j+W_jX_j.$$ This is, in general, not iqual to $$2X_jW_j$$. But it holds if you choose $$X_0=A$$.

You can find details on matrix derivative seraching for "(\frac{dX^2}{dX}) matrix derivative" on SearchOnMath.

Remark: If you follows a decreasing direction, you can arrive at a local minimum or a saddle point to $$\frac{1}{2}\|f(u)\|^2$$. And this can explain some non convergence of Newton's method to a desired root of $$f(u)=0$$.

On the other side, since you are doing the approximation $$u_n\approx u(n)$$, in which $$u(t)$$ is given by equation (1). You can have bad approximation, if $$u_0$$ is far from one root of $$f(u)=0$$, and "lost" the curve you intend to follows.