Maximum-Determinant Optimization Under a Trace Constraint

Why a Fixed Trace Leads to Equal Eigenvalues

A positive-definite matrix can distribute its total size unevenly across directions. Its trace fixes the sum of its eigenvalues, while its determinant is their product. This creates a natural optimization question: if the total amount is fixed, how should it be allocated to make the determinant as large as possible?

The answer is especially useful when a matrix represents a spread or scale across several dimensions. A large determinant corresponds to a large overall volume scale; fixing the trace fixes the total across directions. We will show that the determinant is maximized when no direction receives more than another. The proof also identifies the optimizer uniquely.

The condition \(t>0\) is necessary: every positive-definite matrix has positive eigenvalues and therefore positive trace. For any \(t>0\), the feasible set is nonempty because \((t/p)I_p\) is positive definite and has trace \(t\). Here \(I_p\) denotes the \(p\times p\) identity matrix.

The Maximum-Determinant Theorem

A real symmetric positive-definite matrix has positive eigenvalues \(\lambda_1,\ldots,\lambda_p\). The spectral theorem gives \(\operatorname{tr}(S)=\sum_{i=1}^p\lambda_i\) and \(\det S=\prod_{i=1}^p\lambda_i\). Thus the matrix problem reduces to comparing the product of positive numbers with a fixed sum.

Proof by eigenvalues. Let \(S\in\mathcal{S}_t\), and let its positive eigenvalues be \(\lambda_1,\ldots,\lambda_p\). Their arithmetic mean is \(t/p\). The arithmetic-geometric mean inequality, with equality if and only if all the positive numbers are equal, gives

Since the logarithm is strictly increasing on positive numbers, this implies \(\log\det S\leq p\log(t/p)\). Equality holds exactly when every eigenvalue equals \(t/p\). A real symmetric matrix whose eigenvalues are all \(t/p\) is \((t/p)I_p\), by its orthogonal diagonalization. This matrix is feasible and achieves equality, which proves both attainment and uniqueness. \(\square\)

This argument also explains why the optimization has a solution even though the positive-definite constraint describes an open set: the explicit feasible matrix \((t/p)I_p\) reaches the upper bound. No limiting sequence toward a singular matrix is needed.

A Second Derivation: Stationarity and Strict Concavity

The eigenvalue proof is direct, while matrix calculus provides a useful complementary view. The differential of the log-determinant at a positive-definite \(S\) is \(d(\log\det S)=\operatorname{tr}(S^{-1}dS)\). The trace constraint has differential \(\operatorname{tr}(dS)=0\) along feasible directions.

Introduce a multiplier \(\lambda\) for the constraint and form \(\mathcal{L}(S,\lambda)=\log\det S-\lambda(\operatorname{tr}(S)-t)\). Stationarity with respect to symmetric matrix variations requires \(S^{-1}-\lambda I_p=0\). Thus any stationary point has \(S=\lambda^{-1}I_p\). Its trace constraint forces \(\lambda^{-1}=t/p\), so the only stationary point is \(S^\star\).

To verify that stationarity identifies the global maximum, consider any nonzero symmetric direction \(H\) along a line of positive-definite matrices. Differentiating twice along that line gives

Whenever \(S+uH\) is positive definite, set \(B=(S+uH)^{-1/2}H(S+uH)^{-1/2}\). This is a symmetric matrix, and cyclic invariance of trace gives \(\operatorname{tr}((S+uH)^{-1}H(S+uH)^{-1}H)=\operatorname{tr}(B^2)\). If the eigenvalues of \(B\) are \(b_1,\ldots,b_p\), then \(\operatorname{tr}(B^2)=\sum_i b_i^2>0\): \(B\ne0\) because \(H\ne0\) and the inverse square root is invertible. Therefore the second derivative is strictly negative in every nonzero direction.

Consequently, \(\log\det S\) is strictly concave on the positive-definite matrices. The feasible set is convex: if \(S_0,S_1\) are positive definite with trace \(t\), then for \(0\leq u\leq1\), \((1-u)S_0+uS_1\) is positive definite and has trace \(t\). Along the segment from the stationary point \(S^\star\) to any distinct feasible \(S\), the objective has negative second derivative and derivative zero at \(S^\star\). Its value must therefore decrease strictly away from \(S^\star\). This proves that \(S^\star\) is the unique global maximizer. \(\square\)

Worked Comparisons

Interpretation and a Common Pitfall

The trace \(t\) is the total amount allocated across the \(p\) eigen-directions. At the optimum, each direction receives \(t/p\). In this sense, the solution is an isotropic allocation: no direction is favored, and the matrix is a scalar multiple of the identity. If the matrix describes a covariance scale, for example, fixing its trace fixes the total variance across directions, and the maximizing allocation distributes that amount evenly.

A common mistake is to reason that a larger trace automatically means a larger determinant, or to treat trace and determinant as interchangeable summaries. They are different functions of the eigenvalues: the trace is their sum, and the determinant is their product. Under a fixed sum, unequal positive eigenvalues have a smaller product than equal ones. Another mistake is to conclude that every feasible matrix must be diagonal in the coordinates being used. The theorem imposes a condition on eigenvalues; matrices with off-diagonal entries can be feasible, as the preceding example shows.