Limit of matrix expression $\lim_{B \to A} {(A - B)}^{-1} (A^2 - B^2)$

Question

I'm reviewing linear algebra topics after quite a few years and I'm struggling with this apparently very simple exercise.

Let $A$ be an $n \times n$ matrix.

a. What does it mean to say that $$\lim_{B \to A} {(A - B)}^{-1} (A^2 - B^2)\qquad\text{exists?}$$

b. Does the limit exist when $A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$?

c. When $A = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$?

a. It means that

1. the expression is well-defined and there's a neighborhood $U$ around $A$ such that if $B \in U$ then $A - B$ is invertible, and;

2. that there exists a matrix $M$ such that, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that $|B - A| < \delta$ implies $|{(A - B)}^{-1} (A^2 - B^2) - M| < \varepsilon$, where $|\cdot|$ is the Frobenius norm.

b. In this case $A$ is invertible as $\det A = 1$ and since the determinant is continuous there's a neighborhood around $A$ that contains matrices $B$ so that $A - B$ has a non-zero determinant as well. This should allow us to approach $A$ with a sequence of matrices $B_n$ so that $A - B_n$ is invertible and the limit makes sense.

c. This case seems the same as the previous one. The only difference I see is that for the first matrix, $A^2 = A$, whereas in this case that isn't true. But I don't understand how that has any relevance.

Since I wasn't able to conclude whether the limit exists or not in either case, I tried to evaluate it for the general case. Let $B_n = A - n^{-1} I \to A$ as $n \to \infty$. This sequence belongs to a subset of a neighborhood of $A$ and $A - B_n$ is invertible for $n > 1$. Then we have $${(A - B_n)}^{-1} (A^2 - B_n^2) = nI\left[A^2 - {(A - n^{-1} I)}^2\right] = 2A - n^{-1}I \to 2A$$

This result at least shows that the limit could equal $2A$. But the sequence $B_n$ is only one possible way to approach $A$. We need to show that such a result is independent of how we approach the limit point $A$. In general, we have $B = A - H$, where $|H|$ gets arbitrarily small (say $|H| < \delta$). Then: $${(A - B)}^{-1} (A^2 - B^2) = H^{-1}\left[A^2 - {(A - H)}^2\right] = H^{-1}AH + A - H$$

Now, to prove that the limit is indeed $2A$, we need to bound the following expression with another one that depends on $\delta$ and not on $H$:

$$\begin{split} |H^{-1}AH + A - H - 2A| &= |H^{-1}AH - A - H| \leq\\ &\leq |H^{-1}AH - A| + |H| \leq\\ &\leq |H^{-1}AH - A| + \delta \end{split}$$

And here I don't know how to continue.

Questions:

1. What is the difference between points b and c of the exercise? I could not understand.
2. Is it possible to complete the $\varepsilon-\delta$ proof that the limit is $2A$, or is the above completely wrong?

Answer 1

First notice that the usual definition of $\lim_{x\to a}f(a)$, requiring the domain $D\subset X$ of the function $f$ to be a neighborhood in $X$ of the point $a\in X$, cannot be the one intended here, otherwise the limit in this exercise would never make sense (the subset of matrices $B$ such that $B-A$ is invertible is never a neighborhood of $A$). The intended definition here is a more general one, requiring only $a$ to be adherent to $D$: $$\lim_{x\to a}f(x)=\ell\iff\forall V\in\mathcal V(\ell),\exists U\in\mathcal V(a),f(U\cap D\setminus\{a\})\subset V.$$ This definition can be applied here because "invertible matrices are a dense subset".

With this convention in mind, $\lim_{B \to A} {(A - B)}^{-1} (A^2 - B^2)$ exists iff $\lim_{H\to0} H^{-1} (A^2 -(A-H)^2)$ exists.

But $H^{-1}(A^2 -(A-H)^2)=H^{-1}AH+A-H,$ so the previous limits exist iff the following one does: $$L:=\lim_{H\to0}H^{-1}AH $$ and then we have $\lim_{B \to A} {(A - B)}^{-1} (A^2 - B^2)=L+A.$

Now, the function $g(H):=H^{-1}AH$ (defined on invertible matrices) is homogeneous of degree $1$, i.e. $\forall t\in\Bbb R^*,g(tH)=g(H).$ Therefore, if $L$ exists then for every invertible $H$, $L=\lim_{t\to0}g(tH)=g(H)$. So, the existence of $L$ is equivalent to $g$ being constant (equal to $g(I_n)$), i.e. $$\forall H\text{ invertible},\quad H^{-1}AH=A.$$

This last condition is known to be equivalent to: $A$ is a scalar matrix, and then we have $L=A.$

Answer 2

First off, notice that no matrix $A$ has a neighborhood $U$ such that $A - B$ is invertible for all $B \subset U$. Simply take $B = A - C$ where $C$ is any non-invertible matrix with an arbitrarily small norm. So to define the limit, one must consider only those $B$ such that $A - B$ is invertible.

Regarding your approach, you didn't notice that $H^{-1}AH = A$ when $A$ is the identity, so that the limit is always $2A$. You can also obatain the limit quickly in the following way: If $A$ is the identity, then $AB = BA$ for all matrices $B$. Hence, $A^2 - B^2 = (A - B)(A + B)$. But if $A - B$ is invertible then the expression in the limit reduces to $A + B$. Letting $B$ tend to $A$, we see that the limit is $2A$.

The same trick does not work in (c), because \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} does not commute with all matrices. You would actually get different limits depending on how you approached $A$. For example, taking $B = (1 + t)A$ and letting $t$ tend to $0$, we get the limit $2A$. However, if you let $B$ equal \begin{bmatrix}t & 1\\1 & 2t\end{bmatrix} then (if I have not made a mistake) after some computation you obtain \begin{bmatrix}0 & 3\\\frac{3}{2} & 0\end{bmatrix} in the limit as $t \to 0$.