Show that if $x_{n}$ converges to $x_0$ , then $Ax_0 = y$

Question

Let $A$ be a square matrix of size $N$, which we assume to be invertible. We define the following sequence:

$$x_{n + 1} = x_{n} - A^{T}(Ax_{n} - y), $$

and $x_{0}$ is given.

Show that if $x_{n} \rightarrow x_{0}$, then $Ax = y$.

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I have been stuck on this problem for a while, and I don't really know how to proceed. I thought about using the definition of convergence with $\epsilon-N$, but I haven't gotten anywhere with that.

Can anyone please help me?

Answer 1

Taking limits you get $x_0=x_0-A^{T}(Ax_0-y)$. Hence $A^{T}(Ax_0-y)=0$. Since $A$ is invertible so is $A^{T}$. Hence $Ax_0-y=0$ or $Ax_0=y$.

Answer 2

Hint: If you know that $x_n \to \boldsymbol{\ell}$ say (where $\boldsymbol{\ell}$ is a vector here), then you can send $n$ to infinity on both sides and use limit laws, as well as the fact that matrix multiplication is continuous (i.e. if $x_n\to \boldsymbol{\ell}$, then $Ax_n\to A\boldsymbol{\ell}$; make sure you know why this is true!).

Answer 3

From the given recurrence,

$$Ax_n-y=A^{-T}(x_{n+1}-x_n).$$

Convergence of $x_n$ implies $\exists\,n$

$$\|Ax_n-y\|<\|A^{-T}\|\epsilon$$ for all $\epsilon>0$.