Inequality between induced matrix norms implies equality Let us consider $N_1$ and $N_2$ two induced norms on the space of the square matrices with $n\in\mathbb N^*$ rows and colums.
Assume that $N_1 \leq N_2$.
Is that true that $N_1 = N_2$ ?
[edit] The answer is yes (see Horn and Johnson, Matrix Analysis, 5.6). The proof is based on inequalities with induced norms and the equality is obtained by considering particular rank-one matrices.
 A: For completeness, I present a proof following section 5.6 of Horn and Johnson's Matrix Analysis, which was pointed out by user1551. They prove a stronger result on page 303: if $\|\cdot  \|_\alpha$ and $\|\cdot \|_\beta$ are two induced matrix norms, then the supremum of their ratio is the same regardless of the order:
$$
\sup_{A\ne 0} \frac{\|A  \|_\alpha}{\|A \|_\beta} = \sup_{A\ne 0} \frac{\|A  \|_\beta}{\|A \|_\alpha}
\tag{1}$$
As a special case of (1), if $\|A  \|_\alpha\le \|A \|_\beta$ for all $A$, then also $\|A  \|_\alpha\ge \|A \|_\beta$ for all $A$. 
Proof: Let $\|\cdot \|_a$ and $\|\cdot \|_b$ be the vector norms inducing  $\|\cdot  \|_\alpha$ and $\|\cdot \|_\beta$. Define 
$$R_{ab} = \sup_{x\ne 0} \frac{\|x\|_a}{\|x\|_b},\qquad
R_{ba} = \sup_{x\ne 0} \frac{\|x\|_b}{\|x\|_a}
$$ 
The claim is that both sides of (1) are equal to $R_{ab}R_{ba}$; clearly it suffices to prove this for one of them. For every $x\ne 0$ we have 
$$
\frac{\|Ax\|_a}{\|x\|_a} \le R_{ab}R_{ba} \frac{\|Ax\|_b}{\|x\|_b} 
$$
hence $\|A\|_\alpha\le R_{ab}R_{ba}\|A\|_\beta$. It remains to construct a matrix $A_0$ for which equality is attained. 
Let $y,z$ be two vectors of unit Euclidean norm that attain the suprema $R_{ab}$ and $R_{ba}$, respectively. Let $z_0$ be a normal vector of a supporting hyperplane to the $\|\cdot\|_b$-sphere containing $z$; that is, 


*

*$z_0^*x \le \|x\|_b$ for all $x$, and

*$z_0^*z = \|z\|_b$ 


where the products are inner products. Let $A_0$ be the outer product of $y$ and $z_0$, denoted $yz_0^*$. This is a rank-one matrix whose range is spanned by $y$ and whose kernel is the complement of $z_0$. By construction, $\|A_0z\|_a = \|yz_0^*z\|_a = \|y\|_a \|z\|_b$, hence $\|A_0\|_\alpha\ge R_{ab}R_{ba}\|y\|_b$. 
On the other hand, $\|A_0x\|_b = \|y\|_b |z_0^*x|\le  \|y\|_b\|x\|_b$ for all $x$, which implies $\|A_0\|_\beta\le \|y\|_b$. Thus, 
$$ \|A_0\|_\alpha\ge R_{ab}R_{ba}\|A_0\|_\beta$$
completing the proof.
