Replace matrix approaching zero in a limit with a sequence of integers

Question

Let $Y_k \neq 0$ for all $k \in \mathbb{N}$ be a complex square matrix and let $\lim_{k \to \infty} Y_k = 0$. Let $\lim_{k \to \infty} Y_k / ||Y_k|| = Y \in M_n (\mathbb{C})$. Then $$\lim_{k \to \infty} [(t / ||Y_k||) Y_k] = tY,$$ for all $t \in \mathbb{R}$. How do I prove that for each $t \in \mathbb{R}$ there exists a sequence of integers $\{n_k\}$ such that $$\lim_{k \to \infty} n_k ||Y_k|| = t?$$ From what I am reading, this is implied by the fact that $\lim_{k \to \infty} ||Y_k|| = 0$.

Answer 1

To simplify typing and notation, let $(a(k))_k$ be a sequence of (strictly) positive real numbers which converges to zero. (For our purposes, set for each natural $k$ the value $a(k)=\|Y_k\|>0$.)

Fix $t\ge 0$.

We distinguish two cases:

• $t=0$.

Then we set $n_k$ to be the floor (or the ceil) of $1/a(k)^{1/2}$. It converges to infinity, since the given sequence converges to zero. Also, $n_k$ lies in the interval with extremities $1/a(k)^{1/2}\pm 1$. This sandwich immediately gives $$ \left(\ \frac 1{a(k)^{1/2}}- 1\ \right)a(k) \ \le\ n_ka(k) \ \le\ \left(\ \frac 1{a(k)^{1/2}}+ 1\ \right)a(k) \ . $$ The left most, and the right most expression converge to $t=0$.

• The case $t>0$.

Set $n_k$ to be the floor (or ceil) of $t/a(k)$. The the sandwich is: $$ \left(\ \frac t{a(k)} - 1\ \right)a(k) \ \le\ n_ka(k) \ \le\ \left(\ \frac t{a(k)} + 1\ \right)a(k) \ . $$

Answer 2

You don't actually need to construct the sequence. You only need to prove it exists. You have $\lim_{k \to \infty} ||Y_k|| = 0$. Hence there exists $N_t \in \mathbb{N}$ such that $||Y_k|| \le |t|$ for all $k \ge N_t$. Let $A_k = \{ n \mid n ||Y_k|| \le |t|, n \in \mathbb{N} \}$. $A_k$ is finite by the Archimedean property. Hence $n_k = \mathrm{sgn}(t) \sup A_k$ satisfies $n_k ||Y_k|| \le t$ and $(n_k + 1) ||Y_k|| > t$. Thus $n_k ||Y_k|| \in (t - ||Y_k||, t]$. Since for all $\varepsilon > 0$ there exists $N_\varepsilon \in \mathbb{N}$ such that $||Y_k|| < \varepsilon$ for all $k \ge N_\varepsilon$, $|n_k ||Y_k|| - t| < \varepsilon$. Therefore $\lim_{k \to \infty} n_k ||Y_k|| = t$.