Prove the maximum of the function $\det x$ restrained to the sphere $\sum x_{ij}^2=n$ is $1$

Question

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?

Answer 1

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 $\lambda_i = 1$.

Answer 2

Note that Hadamard's inequality gives $|\det A| \le \|A e_1 \| \cdots \|A e_n\|$. A nice proof based on the Gram Schmidt process (for $\mathbb{R}$) is given here https://www.ma.utexas.edu/users/voloch/Homework/hadamard.pdf. (The same approach works with $\mathbb{C}$.) The result shows that equality is achieved iff the $A e_k$ are orthogonal.

Now note that $\max \{ u_1 \cdots u_n | \sum_k u_k = 1, u_k \ge 0 \}$ has maximum value $ {1 \over n^n}$ which is attained iff $u_k = {1 \over n}$ for all $k$.

Then we see that \begin{eqnarray} \max \{ \|A e_1 \|^2 \cdots \|A e_n\|^2 | \sum_k \|A e_k\|^2 = n \} &\le& \max \{ u_1 \cdots u_n | \sum_k u_k = n, u_k \ge 0 \} \\ &=& n^n \max \{ u_1 \cdots u_n | \sum_k u_k = 1, u_k \ge 0 \} \\ &=& 1 \end{eqnarray} Hence we have $|\det A | \le 1$ whenever $\|A\|_F = n$. Furthermore, we have equality iff $\|A e_k\| = 1$ and the $A e_k$ are orthogonal and $A$ represents a proper rotation (so $\det A = 1$ rather than $-1$).

No Lagrange multipliers were intentionally hurt in the making of this answer.