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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.

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cyc
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  • This is obviously not always possible if $N$ and $\varepsilon$ are not related to each other. E.g. if $N>0$ and $\varepsilon$ are very small $|uv-I|$ and $|vu-I|$ are necessarily close to 1 (and hence not $\le\varepsilon$) when $|u|,|v|\le N$. – user1551 Feb 26 '15 at 11:49
  • @user1551: I forgot to mention the bounds on $N$ and $\varepsilon$. I've added them in now. – cyc Feb 26 '15 at 15:02
  • This is not enough. In order that $|I-uv|\le\varepsilon$, a necessary condition is that $N|u| \ge|uv| \ge1-\varepsilon$. In other words, if $|u|<\frac1N(1-\varepsilon)$, there is no hope to find even a single $v=v_0$ that satisfies your requirement. – user1551 Feb 26 '15 at 16:28
  • I agree. For $v$ to exist, we must have $\frac{1}{N}(1-\varepsilon)\leq||u||\leq N$. But what I want to know is given the existence of $v$, can I connect $u$ to $I$ by a path of matrices with similar properties? – cyc Feb 26 '15 at 16:57

1 Answers1

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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.

user1551
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  • And yes, this is adapted from my answer in the post that the OP linked to. – user1551 Feb 27 '15 at 08:49
  • In your definition of $V_t$, you meant to have $V$ on the right hand side instead of $U$, didn't you? – cyc Feb 27 '15 at 17:14
  • @learner Yes, that's a typo. It's fixed now. Thanks for catching that mistake. – user1551 Feb 27 '15 at 17:28
  • That was a nice proof. I have another question of a similar flavor for which I wonder whether singular value decomposition can be used and that you may have some ideas for: http://math.stackexchange.com/questions/1168565/connecting-an-almost-idempotent-complex-matrix-to-a-diagonal-1-0-matrix – cyc Feb 28 '15 at 00:20