Approaching the limit of an inverse matrix

So I was trying to find the inverse of the $3\times 3$ matrix of ones. I found that its adjoint is the zero matrix and its determinant is $0$ so the entries of the inverse matrix are in an indeterminate form. Suppose that there was a function $f: \mathbb{R}^{6\times1} \rightarrow \mathbb{R}^{3\times3}$. f would change the entries of the first two rows of $3\times3$ matrix of ones and then calculate its inverse. Does there exist a path for which we can take the limit along as f approaches $(1,1,1,1,1,1)$? I already know that the limit doesn't exist in general so now I'm trying to see along what paths does the limit exists and if all such paths yield the same limit.

EDIT: Sorry this was my mistake. The function $f$ is defined as $$f = \frac{adj(A)}{|{A}|}$$ where $adj(A)$ is the Adjoint of A and $|A|$ is determinant of A.

Solution 1:

First of all I think the function you are defining is this: $\mathbf{f}:\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)\in D\subset\mathbb{R}^6\to \begin{bmatrix} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \\ 1 & 1 & 1 \\ \end{bmatrix}^{-1} \in \mathbb{R}^{3\times 3} $

Where $D=\left\{ \mathbf{x} \in \mathbb{R}^6:\exists\begin{bmatrix} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \\ 1 & 1 & 1 \\ \end{bmatrix}^{-1}\right\}$

Now you are asking about: $$\lim_{\mathbf{x}\to (1,1,1,1,1,1)} \mathbf{f}(\mathbf{x})$$ (Where $\mathbb{R}^6$ is equipped with euclidean metric and $\mathbb{R}^{3\times 3}$ with Frobenius metric) First of all I would like to make you notice that two metrics above are both norm-induced metric so in both spaces I can use the respective norms and: $$d(x,y)=||x-y||$$ (For simplicity I'll use the same symbol for the two norms/metrics even if they are different). A theorem(that is pretty advanced!I just had the luck to read it during a numerical calculus course, if you want to know something about it it's a consequence of Eckart-Young theorem)tells us that if $A$ is a non-singular matrix of order $n$(let $\Sigma$ be the set of all the singular matrices of order $n$): $$\min_{S \in \Sigma}(||A-S||)=\frac{1}{||A^{-1}||}$$ Clearly this implies in our particular space that: $$\left\|A-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|\geq \frac{1}{||A^{-1}||} \ \ \ \ \forall A : \exists A^{-1}$$ So: $$\left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|\geq \frac{1}{||\mathbf{f}(\mathbf{x})||} \ \ \ \ \forall \mathbf{x} \in D$$ So: $$||\mathbf{f}(\mathbf{x})|| \geq \left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|^{-1} \forall \mathbf{x} \in D$$ Notice that if our limit exists finite along a path $\mathcal{P}\subseteq D$(such that $(1,1,1,1,1,1)$ is of accumulation for $\mathcal{P}$), then: $$\forall \varepsilon>0 \exists \delta >0 :\forall \mathbf{x} \in \mathcal{P} \ \ \ (||\mathbf{x}-(1,1,1,1,1,1)||<\delta \Rightarrow ||\mathbf{f}(\mathbf{x})-L||<\varepsilon)$$

So if $\mathbf{x}$ is in a sufficiently small ball(let's call it $B$) of center $(1,1,1,1,1,1)$ ,then $||\mathbf{f}(\mathbf{x})||$ is bounded in $B \cap \mathcal{P}$(you can easily prove it using norm subtraction property on $||\mathbf{f}(\mathbf{x})-L||<\varepsilon$). But: $$||\mathbf{f}(\mathbf{x})|| \geq \left\|\mathbf{f}(\mathbf{x})^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|^{-1} \forall \mathbf{x} \in \mathcal{P}\subseteq D$$

So the boundness of $||\mathbf{f}(\mathbf{x})||$ , with $\mathbf{x} \in B \cap \mathcal{P}$ requires $ \left\|\mathbf{f}(x)^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|^{-1}$ with $\mathbf{x} \in B\cap \mathcal{P}$ to have an upper bound. That is to say that $ \left\|\mathbf{f}(\mathbf{\mathbf{x}})^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|$ with $\mathbf{\mathbf{x}} \in B \cap \mathcal{P}$ must have a non-zero lower bound. But: $$\left\|\mathbf{f}(\mathbf{\mathbf{x}})^{-1}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|=\left\|\begin{bmatrix} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \\ 1 & 1 & 1 \\ \end{bmatrix}-\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}\right\|=\sqrt{\sum_{i=1}^6 (x_i-1)^2}=$$ $$=||\mathbf{x}-(1,1,1,1,1,1)||$$ But clearly: $$\inf_{\mathbf{x} \in B \cap \mathcal{P}}(||\mathbf{x}-(1,1,1,1,1,1)||)=0$$ Because $(1,1,1,1,1,1)$ is an accumulation point of $\mathcal{P}$(and also of $B \cap \mathcal{P}$, since $B$ is a neighborhood of $(1,1,1,1,1,1)$). So there is not non-zero lower bound and by consequence the limit cannot exist finite along any path. This concludes the proof.