矩阵范数不等式：基于给定约束求解矩阵差2范数的上下界

Alright, let's work through this question step by step. The short answer is: no, such positive real numbers $\gamma_{min}, \gamma_{max}$ do not generally exist. Here's why, with concrete reasoning and examples:

1. No upper bound $\gamma_{max}$ exists #

We can always construct a matrix $A'$ that satisfies both given constraints while making $\Vert A - A' \Vert_2$ arbitrarily large. Let's break this down:

Suppose $n \geq 2$ (the most common case for $n \times n$ matrices). Pick a non-zero matrix $M$ such that $M x_0' = 0$—for example, the projection matrix onto the orthogonal complement of $x_0'$:
$$M = I - \frac{x_0' (x_0')^T}{\Vert x_0' \Vert_2^2}$$
This matrix has $\Vert M \Vert_2 = 1$, and $M x_0' = 0$ by construction.

Now define $A' = A + kM$, where $k$ is any positive real number we choose. Let's check the constraints:

• $\Vert A x_0 - A' x_0' \Vert_2 = \Vert A x_0 - (A x_0' + k M x_0') \Vert_2 = \Vert A(x_0 - x_0') \Vert_2 \leq \Vert A \Vert_2 \beta_2$. If $\beta_1 \geq \Vert A \Vert_2 \beta_2$, this satisfies the first constraint. Even if $\beta_1 < \Vert A \Vert_2 \beta_2$, we can adjust $k$ or tweak $M$ slightly to still meet the constraint. The core point remains: $\Vert A - A' \Vert_2 = k \Vert M \Vert_2 = k$, which can be made as large as we want.

If $x_0' = 0$, we can set $A'$ to any matrix (since $A' x_0' = 0$), making $\Vert A - A' \Vert_2$ arbitrarily large while satisfying $\Vert A x_0 - 0 \Vert_2 \leq \beta_1$.

2. No positive lower bound $\gamma_{min}$ exists #

We can also make $\Vert A - A' \Vert_2$ arbitrarily close to 0, violating the requirement that $\gamma_{min} > 0$:

• Simply take $A' = A + \epsilon I$, where $\epsilon$ is a tiny positive number. For small enough $\epsilon$, $\Vert A x_0 - A' x_0' \Vert_2 = \Vert A(x_0 - x_0') - \epsilon x_0' \Vert_2 \leq \Vert A \Vert_2 \beta_2 + \epsilon \Vert x_0' \Vert_2$, which will be $\leq \beta_1$ once $\epsilon$ is small enough. The second constraint $\Vert x_0 - x_0' \Vert_2 \leq \beta_2$ is unchanged.
• If we set $\epsilon = 0$, then $A' = A$, so $\Vert A - A' \Vert_2 = 0$, which is strictly less than any positive $\gamma_{min}$.

Edge Cases #

The only scenario where bounds might seem possible is when $n=1$ (scalars) and $x_0' \neq 0$, but even then:

• We can still make $\Vert A - A' \Vert_2$ arbitrarily close to 0 (by taking $\epsilon \to 0$), so no positive $\gamma_{min}$ exists.
• For the upper bound, while we can derive a finite limit if $\beta_1$ and $\beta_2$ are fixed, this only applies to 1-dimensional scalars, not general $n \times n$ matrices.

内容的提问来源于stack exchange，提问作者Srinath