Any relation between the singular values of ${\bf A}$ and ${\bf I} - {\bf A}$ Is there any relation between the singular values of ${\bf A}$ and ${\bf I} - {\bf A}$? When $\bf A$ is Hermitian, then the singular values of $\bf A$ is just it eigenvalues, and ${\bf I} - {\bf A}$ has its singular values being $1$ minus those eigenvalues. Is there any relation when $\bf A$ is not Hermitian? Thanks!

For readers of this post, see also If the absolute value of every eigenvalue of a matrix is smaller than 1, is the maximum singular value smaller than 1?.
 A: The singular values of $I-A$ are not determined by those of $A$. To illustrate, consider
$$
A_1=\pmatrix{1&-1\\ 1&1},\quad A_2=\pmatrix{-1&1\\ -1&-1},\quad A_3=\pmatrix{1&1\\ 1&-1}.
$$
The singular values of these three matrices are all equal to $\sqrt{2}$, but the singular values of $I-A_1$ are $1,1$, the singular values of $I-A_2$ are $\sqrt{5},\sqrt{5}$ and the singular values of $I-A_3$ are $\sqrt{2}+1,\ \sqrt{2}-1$.
One can, however, relate the singular values of $A$ and $I-A$ by inequalities. See the other answer here by Qiaochu Yuan, for instance.
A: In general, the singular values of two matrices $M, N$ of the same dimension are related to the singular values of $M + N$ by various inequalities, for example the Weyl inequalities
$$\sigma_{k+\ell+1}(M + N) \le \sigma_{k+1}(M) + \sigma_{\ell+1}(N).$$
See this blog post for a proof. We can get various constraints relating the singular values of $A$ and $I - A$ by careful plugging in. For example, with $M = A, N = I - A$, we get
$$\sigma_{k+\ell+1}(I) \le \sigma_{k+1}(A) + \sigma_{\ell+1}(I - A)$$
where $\sigma_{k+\ell+1}(I) = 1$ if $k+\ell+1 \le n$ and $0$ otherwise. This tells us that for $k+\ell+1 \le n$ the singular values $\sigma_{k+1}(A)$ and $\sigma_{\ell+1}(I - A)$ can't both be too small; for example, one of them must be at least $\frac{1}{2}$. (You can think of this as a quantitative version of the observation that the ranks of $A$ and $I - A$ must add up to at least $n$, where all of the matrices here are $n \times n$.) 
With $M = I, N = -A$, we get
$$\sigma_{k+\ell+1}(I - A) \le \sigma_{k+1}(I) + \sigma_{\ell+1}(A).$$
Since the singular values are in decreasing order we get the strongest version of this claim by setting $k = 0$, which gives
$$\sigma_{\ell+1}(I - A) \le 1 + \sigma_{\ell+1}(A).$$
This tells us that the singular values of $I - A$ can't be too much larger than the singular values of $A$.
Finally, with $M = I, N = A - I$, we get
$$\sigma_{k+\ell+1}(A) \le \sigma_{k+1}(I) + \sigma_{\ell+1}(I - A).$$
As before the strongest version of this claim comes from setting $k = 0$ and gives
$$\sigma_{\ell+1}(A) \le 1 + \sigma_{\ell+1}(I - A)$$
which says that the singular values of $A$ can't be too much larger than the singular values of $I - A$. This is the same as the previous claim but with the roles of $A$ and $I - A$ switched. 
