If $A$ is matrix in $\mathcal M_n(\Bbb R)$ we define the norm $$\Vert A\Vert = \sqrt{\operatorname{Tr}(A^TA)}$$
Now I'm wondering is it true that over the set of special linear matrices $M$ i.e. such that $\det(M)=1$ the matrices with minimal norm are the special orthogonal matrices? In other words if $A$ such that $\det(A)=1$ and $\Vert A\Vert\le \Vert M\Vert$ for all special linear matrix $M$ then $A$ is special orthogonal i.e. $A^TA=I_n$.
I searched in internet for a proof but I didn't find any thing.
Note that in terms of the singular values $\sigma_i$ of $A$, we have $$ \|A\| = \sqrt{\sigma_1^2 + \cdots + \sigma_n^2}. $$ On the other hand, we have $|\det(A)| = \sigma_1 \cdots \sigma_n$. So, let $\lambda_i = \sigma_i^2$ (in fact, $\lambda_i$ is also the $i$th eigenvalue of $A^TA$). $A$ will be of minimal norm if and only if $\lambda_1,\dots,\lambda_n$ are such that $$ \lambda_1 + \cdots + \lambda_n = \min \{\lambda_1 + \cdots + \lambda_n : \lambda_i \geq 0 \text{ for all }i, \lambda_1 \cdots \lambda_n = 1\}. $$ It is well-known that this holds exactly when $\lambda_i = 1$ for all $i$, which is to say that $\sigma_i = 1$. This can be shown in a straightforward way with the AM-GM inequality, Lagrange multipliers, or Jensen's inequality.
Now if $\sigma_i = 1$, then the singular value decomposition yields $$ A = U I V^T = UV^T. $$ So indeed: a matrix $A$ will minimize $\|A\|$ subject to the constraint that $\det(A) = 1$ if and only if $A$ is orthogonal.