# If $\lVert \cdot \rVert \leq \lVert \cdot \rVert'$ are two norms on $\mathbb{R}^n$, what would be the relation of the corresponding operator norms?

Let $$T : \mathbb{R}^n \to \mathbb{R}^n$$ be a linear map.

Since $$\mathbb{R}^n$$ is finite-dimensional, all norms are equivalent on it. Thus, all operator norms of $$T$$ must be equivalent as well.

However, I would like to know more details. That is, if $$\lVert \cdot \rVert$$ and $$\lVert \cdot \rVert'$$ are two norms on $$\mathbb{R}^n$$ such that $$$$\lVert v \rVert < \lVert v \rVert'$$$$ except for $$v=0$$, then is there any definite relation between $$\lVert T \rVert_{op}$$ and $$\lVert T \rVert'_{op}$$?

Here, $$\lVert T \rVert_{op}:=\sup_{v \neq 0} \frac{\lVert Tv \rVert}{\lVert v \rVert}$$ and similarly for $$\lVert T \rVert'_{op}$$.

The fraction appearing in the definition of the operator norm prevents me from making a conclusion. Could anyone please help me?

• You can avoid fractions $\|T\|=\sup_{\|x\|\le 1}\|Tx\|.$ Commented May 29, 2023 at 13:32

Suppose $$\frac1c ||\cdot || \le || \cdot ||' \le c ||\cdot||$$ with $$c > 0$$. Then $$\frac{||Tv||'}{||v||'} \le \frac{c||Tv||}{\frac1c ||v||} \le c^2 \frac{||Tv||}{||v||}.$$ Taking supremum, $$||T||_{\text{op}}' \le c^2 ||T||_{\text{op}}$$. By symmetry, $$\frac1{c^2} ||T||_{\text{op}} \le ||T||_{\text{op}}' \le c^2 ||T||_{\text{op}}$$.