Question

Let $s_1,s_2: \mathbb{R}^{2\times 2} \mapsto \mathbb{R}_+$, $s_{1}\left(\cdot\right)\ge s_{2}\left(\cdot\right)\ge 0$, be the singular values of a $2\times2$ matrix. Is it true that $$\left|s_{1}\left(M+N\right)-s_{1}\left(N\right)\right|+\left|s_{2}\left(M+N\right)-s_{2}\left(N\right)\right|\leq s_{1}\left(M\right)+s_{2}\left(M\right)$$ for any two $2\times2$ real matrices $M$ and $N$?

Thanks in advance for any helpful answers.

Answer 1

Here is a more general result.

Let $A$ and $B$ be arbitrary $n \times n$ complex matrices. Then, we have the weak-majorization:

$$ |s(A) - s(B)|\quad \prec_w\quad s(A-B)$$

This result implies your alleged inequality as a special case.

The above result follows from a famous theorem of Lidskii, which states that for Hermitian matrices $A$ and $B$,

$$ \lambda^\downarrow(A) - \lambda^\downarrow(B) \prec \lambda(A-B),$$ where $\lambda^\downarrow(A)$ lists eigenvalues of $A$ is decreasing order (notice that here the majorization is strict)

For more details, see for example, Exercise IV.3.1 in Matrix Analysis by R. Bhatia.

Alternatively, you can have a look (for the singular value majorization result) at Theorem 3.4.5 in Topics in Matrix Analysis by Horn and Johnson.