Path-connectedness of a certain set of invertible complex matrices We know that $GL_n(\mathbb{C})$ is connected. I am considering the following variant that has been bothering me for quite some time. 
Let $u\in M_n(\mathbb{C})$ be such that $||u||\leq N$, $||uv-I||<\varepsilon$, $||vu-I||<\varepsilon$ for some $v\in M_n(\mathbb{C})$ with $||v||\leq N$ and some $\varepsilon>0$. Is there a continuous path $(u_t)_{t\in[0,1]}$ in $M_n(\mathbb{C})$ such that $u_0=u$, $u_1=I$, and for each $t$, $||u_t||\leq N$, $||u_t v_t-I||<\varepsilon$ and $||v_t u_t-I||<\varepsilon$ for some $v_t\in M_n(\mathbb{C})$ with $||v_t||\leq N$? 
Added remark: I consider only $N\geq 1$ and $\varepsilon<\frac{1}{4}$.
The norm I'm using is the matrix 2-norm. Such matrices are in fact invertible but the point is the control on the norm.
In the conditions for $u_t$ and $v_t$, I can settle for something like $4\varepsilon$ and $4N^2$ if that helps.
In this post, there seems to be a few different ways to show that $GL_n(\mathbb{C})$ is connected. I wonder whether any of them can be adapted for my purpose.
 A: For the operator norm (spectral norm), the answer is affirmative and an explicit construction is possible.
Proposition. Let $\|\cdot\|$ denotes the operator norm. Suppose $0<\varepsilon<1\le N$ and $u\in GL_n(\mathbb C)$. If there exists some $w\in GL_n(C)$ such that $\|w\|\le N$ and $\|uw-I\|\le\varepsilon$, then there exists a continuous path $\{(u_t,v_t): 0\le t\le1\}$ in $GL_n(\mathbb C)^2$ such that
\begin{cases}
u_0=u,\\
(u_1,v_1)=(I,I),\\
\|u_t\|, \|v_t\|\le N\ \text{ for all } t,\\
\|u_tv_t-I\|=\|v_tu_t-I\|\le\varepsilon\ \text{ for all } t.\\
\end{cases}
(Note, however, that $v_0$ is not necessarily $w$, but possibly some other matrix.)
Proof. Let $\mathbf y$ and $\mathbf x$ be respectively the left and right unit singular vector of $u$ for the smallest singular value $\sigma_n$. By the given assumption, we have
$$
1-N\sigma_n
\le 1-\sigma_n\|\mathbf x^\ast w\|_2
\le\|\sigma_n \mathbf x^\ast w-\mathbf y^\ast\|_2
=\|\mathbf y^\ast(uw-I)\|_2
\le\varepsilon.
$$
It follows that $\sigma_n$, and in turn all singular values of $u$, lie inside the interval $J=\left[\frac1N(1-\varepsilon),\,N\right]$. Let $f:J\to J$ be the continuous function defined by
$$
f(x)=\frac1{\max(x,\frac1N)}.
$$
By singular value decomposition and the fact that every unitary matrix can be unitarily diagonalised, $u=(UDU^\ast)\Sigma (V\tilde{D}V^\ast)$ for some unitary matrices $U,\,V$, some unit diagonal matrices $D=\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n}),\,\tilde{D}=\operatorname{diag}(e^{i\phi_1},\ldots,e^{i\phi_n})$ and some singular value matrix $\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)$. Let $s_i:[0,1]\to J$ be any continuous path such that $s_i(0)=\sigma_i$ and $s_i(1)=1$. Define
\begin{align}
U_t&=U\operatorname{diag}(e^{i(1-t)\theta_1},\ldots,e^{i(1-t)\theta_n})U^\ast,\\
V_t&=V\operatorname{diag}(e^{i(1-t)\phi_1},\ldots,e^{i(1-t)\phi_n})V^\ast,\\
u_t&=U_t\pmatrix{s_1(t)\\ &s_2(t)\\ &&\ddots\\ &&&s_n(t)}V_t^\ast,\\
v_t&=V_t\pmatrix{f(s_1(t))\\ &f(s_2(t))\\ &&\ddots\\ &&&f(s_n(t))}U_t^\ast.
\end{align}
Then all requirements in the proposition are satisfied.
