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.