If $A$ is a symmetric matrix, then $\det(A) \leq \prod\limits_{i = 1}^d a_{ii}.$

determinantlinear algebramatricesoptimizationsymmetric matrices

Prove or provide a counterexample. If $A = (a_{ij})$ is a symmetric matrix, then $\det(A) \leq \prod\limits_{i = 1}^d a_{ii}.$

The result is obviously true for diagonal matrices and here is a proof for positive semidefinite matrices.

Proof when $A$ is also positive semidefinite. If the smallest eigenvalue of $A$ is zero, then $\det(A) = 0$ and we only need to show that $\prod\limits_{i = 1}^d a_{ii} \geq 0,$ which follows from the fact that $a_{ii} = c_i^\intercal A c_i,$ where $c_i$ is the $i$th canonical vector.

Suppose that the smallest eigenvalue of $A$ is positive, this implies that $A$ is positive definite. We can orthogonally diagonalise
$$
A = P\Lambda P^\intercal = \lambda_1 x_{(1)} x_{(1)}^\intercal + \ldots + \lambda_d x_{(d)}x_{(d)}^\intercal,
$$
where $P = \left[x_{(1)}, \ldots, x_{(d)}\right]$ is an orthogonal matrix and $\lambda_1 \geq \ldots \geq \lambda_d > 0.$ Write $P^\intercal = [x_1, \ldots, x_d],$ i.e. the $x_i$ are the rows of $P.$ Then
$$
a_{ii} = \sum_{j = 1}^d \lambda_j x_{ij}^2 = u(x_i),
$$
where $u$ is the function $u(\xi_1, \ldots, \xi_d) = \lambda_1 \xi_1^2 + \ldots + \lambda_d \xi_d^2.$ Therefore, we have
$$
\prod_{i = 1}^d a_{ii} = \prod_{i = 1}^d u(x_i).
$$

We now show that
$$
(\min u(x_1) \cdots u(x_d) \quad \text{subject to} \quad [x_1, \ldots, x_d] \quad \text{is orthogonal}) = \lambda_1 \cdots \lambda_d.
$$
If this is true, we are done.

  1. We first minimise over the last column $x_d$ subject only to the contraint that $\|x_d\|^2 = 1.$ It is obvious that $u(x_d) \geq \lambda_d$ with equality when $x_d = c_d$ (the $d$th canonical vector). We can substitute this partial result, and our problem now becomes
    $$
    (\min u(x_1) \cdots u(x_{d-1}) u(c_d) \quad \text{subject to} \quad [x_1, \ldots, x_{d-1}, c_d] \quad \text{is orthogonal}) = \lambda_1 \cdots \lambda_d.
    $$
  2. Since $u(c_d) = \lambda_d > 0,$ we may cancel it, and clearly the condition "$[x_1, \ldots, x_{d-1}, c_d]$ is orthogonal" is equivalent to "$[x_1, \ldots, x_{d-1}]$ is orthogonal" (in $\mathbf{R}^{d-1}$ where we assume all last entried of $x_1, \ldots, x_{d-1}$ are zero).
  3. Therefore, we have the same problem as before but one dimension less, and by an easy induction, we are done. QED

Best Answer

This is not true for a general symmetric matrix. For instance,

$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} $$

has $\det A = 1 > \prod_i a_{ii} = 0$.

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