Do we have for all $M \in SL_n(\Bbb K)$, $\lVert M \rVert \geq 1$ when $\lVert \cdot \rVert$ is a matrix norm?

Question

Let be $\lVert \cdot \rVert$ a matrix norm (submultiplicative).

Do we have for all matrices of determinant 1, the following lower bound:

$$\lVert M \rVert \geq 1$$

I'm very confused and could not find any counterexample and I find this statement very fishy, I tried to experiment with:

\begin{bmatrix} 1& x \\ 0& 1 \end{bmatrix}

But, its Frobenius norm cannot be small enough.

Answer 1

The norm of a matrix is larger than its eigenvalues. The determinant is the product of its eigenvalues. So, if the determinant of $M$ is $1$, there must be at least one eigenvalue $\lambda$ such that $|\lambda| \ge 1$, which implies $$\|M\| \ge |\lambda| \ge 1,$$ regardless of the matrix norm used.

Answer 2

If $\lVert M\rVert<1$, then, if $B$ is the closed unit ball, the volume of $M(B)$ will be smaller than the valume of $B$. But that cannot happen because, sense, $\det M=1$, the map $v\mapsto M.v$ must preserve volumes.

Answer 3

If $\| \cdot \|$ is a matrix norm then $\rho(A) \leq \| A\|$ and $\rho(A) = 1$

Answer 4

Well, in fact, I just got it.

$SL_n(\Bbb K)$ is closed.

If I suppose that $\lVert M \rVert < 1$, then: $M^n \in SL_n(\Bbb K)$ for all $n$, and then: $M^n \to 0$ for $\lVert \cdot \rVert$, thus: we derive a contradiction.

Which yields the result.