SVD (or polar decomposition) of a matrix plus a constant Say I have a matrix $\hat M$ with a singular value decomposition (SVD) $\hat M=\hat U\hat D\hat V ^\dagger$. Given this SVD, is there a simple way to get the SVD of $(\hat{M}-z\hat{1})$, for $z\in\mathbb{C}$ some constant (here $\hat 1$ denotes the identity matrix)?
Equivalently, if I have a polar decomposition $\hat{M}=\hat U\hat P$, is there a simple way to get the polar decomposition of $(\hat{M}-z\hat{1})$?
I'm actually only interested in getting the product $\hat{U}\hat{V}^\dagger$ in the case of the SVD, or the unitary part $\hat{U}$ in the polar decomposition, so if there's a method that doesn't get me the singular values $\hat{D}$ or positive-definite part $\hat P$, that's fine too.
 A: I assume that you have checked (because I have not!) the equivalence of the polar decomposition version of your question with the SVD version of it, and so I will go straight to the polar decomposition (PD) version.
By 'simple way' you obviously mean a method that requires less computation than the brute force method, i.e. computing the polar decomposition of $M-zI$ from the grounds up (i.e. without knowing anything about the polar decomposition of $M$).
If there was a general trick for computing the 'translated polar decomposition' that works independent of dimension, it would need to be applicable also in the case where $M$ is just a complex number (i.e. a $1\times1$ matrix). So, let's check if such a method is known to exists in that special case.
The PD of a complex number $M$ is
$$M=r\cdot {\rm e}^{{\rm i}\phi}$$
where
$$r=\sqrt{\Re(M)^2 + \Im(M)^2} \qquad , \qquad \phi={\rm arctan2}(\Re(M), \Im(M))$$
So much for the brute force computation of the PD of $M$. Then your question means for our special case: is there a simpler way to compute the PD of 
$$M-z=r^\prime \cdot {\rm e}^{{\rm i}\phi^\prime}$$ 
than to simply insert it into the above formulae again, i.e.
$$r^\prime =\sqrt{\Re^2(M-z) + \Im^2(M-z)} \qquad , \qquad \phi^\prime={\rm arctan2}(\Re(M-z), \Im(M-z))$$
You can quickly visualize that the complex zero, the point $M$ and the point $M-z$ generally form a non-right triangle in the complex plane. All the theorems I know about non-right triangles make use of trigonometric functions and hence, can be deduced by referring to equivalent right triangles (e.g. by using an altitude). 
Therefore, it is rather unlikely that there is a more clever way of computing $r^\prime$ and $\phi^\prime$ than by going through the real and imaginary parts and using the above formula.
Just take the radius (which is certainly algebraically easier to handle than the transcendent arctangent function) for example. We can write
$$(r^\prime)^2=\Re^2(M) -2\Re(M)\Re(z)+\Re^2(z)+\Im^2(M) -2\Im(M)\Im(z)+\Im^2(z)$$
leading to
$$r^\prime=\sqrt{r^2 - (2r{\rm cos}\phi)\Re(z)-(2r{\rm sin}\phi)\Im(z)+|z|^2}$$
So you definitely can express the new polar radius (corresponding to the symmetric part of the general polar decomposition) by the old one. But is this computation simpler than the straightforward one? Clearly no! The original formula required 3 additions, 2 multiplications and a square root. The 'simplified' version requires 3 additions, 8 multiplications, and a square root. (Remark: well, actually I admit that this argument is not very robust because there could be an efficiency gain in higher dimension but not in $C$)
And presumably, the situation does not get easier if you have multiple angle sines/cosines (unitary part of the PD) and multiple scalings (symmetric part of the PD)
Of course, this is not a strict proof. There could be new undiscovered 21'st century trigonometry that simplifies the computation in the scalar complex case and is also applicable to higher dimensional matrices. And lastly, there could be special dimension numbers, where simplifications exist, although there is no such method for the scalar complex case.
But I think this is pretty unlikely, at least if I have not made a mistake in my argument.
