Given a $3 \times 3$ matrix in which the sum of the squares of all the elements is not more than one, what is the maximal determinant of this matrix?
I have only one idea that equal elements are located on the main diagonal, and the remaining elements are zero.
Best Answer
This determinant measures the volume of the parallelepiped built on the column vectors. If you keep the vector lengths constant, which keeps the sum of squares constant, the volume is maximized when the vectors are orthogonal.
Now let the lengths be $a,b,c$, you want to maximize
$$V=abc$$ under the constraint $$a^2+b^2+c^2=1.$$
By Lagrangian multipliers,
$$abc+\lambda(a^2+b^2+c^2-1)$$
$$\begin{cases}bc+2\lambda a&=0, \\ac+2\lambda b&=0, \\ab+2\lambda c&=0, \\a^2+b^2+c^2-1&=0\end{cases}$$
Eliminating $\lambda$ from the three first equations, you get $a^2=b^2=c^2$, and using the last, $a^2=b^2=c^2=\dfrac13$. Hence
$$\Delta=\frac1{\sqrt{27}}.$$
Direct vector proof:
Let $\vec a,\vec b,\vec c$ be the column vectors of the matrix. The determinant is the mixed product
$$\Delta=\vec a\cdot(\vec b\times\vec c).$$
Using a Lagrangian multiplier, we maximize
$$\vec a\cdot(\vec b\times\vec c)+\lambda(\vec a^2+\vec b^2+\vec c^2-1).$$
Taking the gradient yields
$$\begin{cases}\vec b\times\vec c+2\lambda\vec a=0, \\\vec c\times\vec a+2\lambda\vec b=0, \\\vec a\times\vec b+2\lambda\vec c=0, \\\vec a^2+\vec b^2+\vec c^2-1=0. \end{cases}$$
Then applying dot products on the first equations, we get
$$\lambda\vec a\vec b=\lambda\vec b\vec c=\lambda\vec c\vec a=0,$$ which shows that the vectors must be orthogonal, and we are back to the above scalar problem.