Prove the maximum of the function $\det x$ restrained to the sphere $\sum x_{ij}^2=n$ is $1$ Let $M(n\times n)=\mathbb{R}^{n^2}$. Prove the maximum of $f:\mathbb{R}^{n^2}\to\mathbb{R}$, given by $f(x)=\det x$, restrained by the sphere $\sum_{i,j} x_{ij}^2=n$, is reached in an orthogonal matrix, so its value is $1$.
Attempt: define $\varphi=\sum_{i,j} x_{ij}^2=n$, so $\nabla\varphi(x)=2[x_{ij}]$
$\nabla f(x)=(-1)^{i+j}X_{ij}$, with $X_{ij}$ the $ij$-minor of $x$.
Using the Lagrange Multipliers, with $M=\varphi^{-1}(n)$, the maximum of $f|_M$ is given when $(-1)^{i+j}X_{ij}=2\lambda x_{ij}\Rightarrow(-1)^{i+j}X_{ij}x_{ij}=2\lambda x_{ij}^2$. 
I sum in $i$ and $j$ to get: $n\det x=2\lambda\sum_{i,j} x_{ij}^2=2\lambda n\Rightarrow\det x = 2\lambda$. 
I couldn't go further, specially proving that the matrix is orthogonal. Any help?

The sum $\sum x_{ij}^2 = \operatorname{tr} x^t x$, so the eigenvalues $\lambda_1, \dots, \lambda_n$ of the diagonalizable matrix $x^t x$ satisfy $\lambda_1 + \cdots + \lambda_n = n$. Furthermore, each $\lambda_i \geq 0$, since any eigenvector $v_i$ belonging to $\lambda_i$ has 
$$\lambda \langle{v_i, v_i}\rangle 
  = \langle{x^t x v_i, v_i}\rangle 
  = \langle{x v_i, x v_i}\rangle
  \geq 0.$$
Now show that $(\det x)^2 = \det x^t x = \lambda_1 \cdots \lambda_n$ is maximized, subject to the constraints above on the $\lambda_i$, when all $\lambda_i = 1$.
