If $A,H\in GL_{n}(\mathbb{R})$, What happen for $H A H^{-1}$, if $h(i,j)\longrightarrow‎ 0$?

Question

If $A$ and $H$ is a two $n\times n$ matrixs, such that $\det(A)\neq 0$ and $ \det(H)\neq 0$, what happen for $H A H^{-1}$, if $h(i,j)‎\longrightarrow‎ 0 $? Is $H A H^{-1}‎\cong A$?$ \ \ \ (1\leq i,j\leq n).$ Any idea can be helpful.thanks

Answer 1

This is a wrong question because the limit of $HAH^{-1}$ in general may not exist. The right question should be: provided that $\lim_{H\to 0}HAH^{-1}$ exists, is it necessarily similar to $A$?

The answer, unfortunately, is no. Here is a counterexample: $$ HAH^{-1} = \pmatrix{x^2&0\\ 0&x} \pmatrix{1&1\\ 0&1} \pmatrix{\frac1{x^2}&0\\ 0&\frac1x} = \pmatrix{1&x\\ 0&1}. $$

If you mean that a certain entry of $H$ approaches to zero while the others are kept constant, the answer is still no. We only need to modify the above counterexample by a little bit: $$ HAH^{-1} = \pmatrix{x&0\\ 0&1} \pmatrix{1&1\\ 0&1} \pmatrix{\frac1x&0\\ 0&1} = \pmatrix{1&x\\ 0&1}. $$

Edit. Note that the $A$ above is not diagonalisable. In general, the similarity orbit of a real or complex matrix $A$ is closed if and only if $A$ is diagonalisable over $\mathbb C$.

Answer 2

I assume that $h(i,j)$ denotes the coefficients of $H$.

In the case $n=1$ the product $HAH^{-1}$ is a product of numbers, in which $H$ converges to $0$. Then obviously the product converges to $A$.

In the case $n>1$ one first observes, that matrix multiplication is a continous operation considering for example the euclidean metric on $\mathbb{R}^{n\times n}$ (or any $p$-metric, $p\in\mathbb{N}\cup\infty$). Convergence with respect to that metric is the same as coefficient-wise concergence.

By the theorem of Cayley-Hamilton $H^n+a_{n-1}H^{n-1}+\ldots +a_0=0$ for some real numbers $a_k$, where $a_0\neq 0$ by assumption (up to sign $a_0$ is the determinant of $H$). Hence

$H^{-1}=-\frac{1}{a_0}(H^{n-1}+a_{n-1}H^{n-2}+\ldots +a_1)$.

Consequently

$\begin{array}{rcl}\lim_{H\rightarrow 0}(HAH^{-1})&=&(\lim_{H\rightarrow 0}H)A(\lim_{H\rightarrow 0}-\frac{1}{a_0}(H^{n-1}+a_{n-1}H^{n-2}+\ldots +a_1))\\ &=&0\;A\;(-\frac{a_1}{a_0}E)=0, \end{array}$

where $E$ is the unit matrix.