Task is to find maximal possible determinant value for 2×2 and 3×3 matrices given following constraint:
$$\sum_{i,j=1}^na_{ij}^2 \le 1$$
I was able to come up with solution, but I received the test result, where I scored zero for this task. I still hasn't got a chance to take a look at my examined paper, but for now I just want to find whether or not my solution was actually true.
Solution
First trivial idea is that optimum would be reached in the point of exact equality $\sum_{i,j=1}^na_{ij}^2 = 1$.
Then, terms which participate in det calculation with minus sign, must be negative or zero.
$$\left( \begin{matrix} \frac 1 {\sqrt 2} & 0 \\
0 & \frac 1 {\sqrt 2} \end{matrix}\right)$$
$$\left( \begin{matrix} \frac 1 {\sqrt 3} & 0 & 0 \\
0 & \frac 1 {\sqrt 3} & 0 \\
0 & 0 & \frac 1 {\sqrt 3} \end{matrix}\right)$$
So, I came up with these matrices and respective det values are $\frac 1 2$ and $\frac 1 {3\sqrt 3}$.
Is that correct?
Thanks!
Best Answer
Here is a geometric argument that I would give:
It is always possible to decompose any matrix into a product of an orthogonal matrix and an upper triangular matrix. Orthogonal transformation is isometry, so $\sum\limits_{i,j=1}^{n}a_{ij}^{2}$ is preserved on the upper triangular matrix. Orthogonal matrix have determinant 1 or -1, so no effects on absolute value of determinant. Hence we need to look only at determinant of upper triangular matrix with the same constraint, but determinant there are only affected by the diagonal.