# How to maximize $\det(W^*DW)$, where $D$ is real diagonal?

Suppose $$D$$ is a diagonal matrix, $$D=\textrm{diag}(d_1, d_2, \cdots, d_n)$$ with $$d_1> d_2 > \cdots > d_n\ge 0$$. Let $$W=[\pmb{w_1}, \pmb{w_2}, \cdots, \pmb{w_m} ]$$ be a $$n\times m$$ matrix ($$m\le n$$) whose columns are each of unit norm, i.e. $$\Vert\pmb{w_i}\Vert=1$$. What $$W$$ maximizes $$\det(W^*DW)$$?

For $$m=1$$, clearly any eigenvector of $$D$$ corresponding to eigenvalue $$d_1$$ will do. E.g. $$W=e_1\triangleq [1\,0\cdots0]^T$$, and $$\det(W^*DW)=d_1$$.

What about $$m\ge2$$? Any $$W$$ whose columns consist of normalized eigenvectors of $$D$$ corresponding to the $$m$$ largest eigenvalues of $$D$$? How to prove it?

Note $$\det(W^*DW)\geq 0$$ since $$W^*DW\succeq \mathbf 0$$ and that all $$m$$ columns of $$W$$ should be selected to be linearly independent, otherwise we get a determinant of zero which can only be a maximum in the trivial case of $$d_n=0$$ and $$n=m$$ (and any choice of $$W$$ is a maximum in such a case).

Since $$W$$ is injective, run 'thin' QR factorization $$W=QR$$ there $$R$$ is invertible and $$Q$$ is tall and skinny. (Recall this means that $$Q$$ has orthonormal columns and $$R$$ has positive diagonal elements.)

$$\det\big(W^*DW\big)= \det\big(R^*Q^*DQR\big)= \det\big(R^*\big)\cdot \det\big(Q^*DQ\big)\cdot \det\big(R\big)$$
$$\leq \det\big(Q^*DQ\big)$$
$$\leq \prod_{k=1}^m d_k$$

where the inequalities are
(i) $$\det\big(R\big) \leq 1$$ since $$\big \Vert \mathbf w_i\big \Vert_2 =1$$ implies each column of $$R$$ has length one, hence its diagonal elements (which are positive) are at most one but $$R$$ is triangular so its determinant is given by the product of its diagonal.
(ii) recognizing that the determinant is the product of the eigenvalues of a matrix, apply Cauchy Interlacing to $$Q^*DQ$$ to see that its kth largest eigenvalue observes $$0\leq \lambda_k \leq d_k$$ and multiply over the bound.

Conclude:
in the maximizing case $$R = I$$ and so $$W$$ has orthonormal columns which should be given by standard basis vectors

• Cool. I appreciate it. One minor clarification: Obviously $W$ is not unique. e.g. any $\pmb{w_i}=\alpha\cdot e_i,$ with $|\alpha|=1$ will do. Jul 6 at 6:23
• How does eigenvalue interlacing apply to $Q^*DQ$? Usually it applies to a principal submatrix. Jul 11 at 5:33
• @copper.hat Cauchy Interlacing merely requires $Q^*Q = I_m$. What you are stating is a special case of Interlacing, though one formulation implies the other. E.g. for your statement select $V\in U_n(\mathbb C)$ (i.e. $V^{-1}=V^*$) such that the first $m$ columns agree with $Q$, i.e. $Q=VS$ where $S$ is tall and skinny and $\mathbf s_j = \mathbf e_j$ (std basis vector). So $Q^*DQ=S^*V^*DVS=S^*\big(V^*DV\big)S$ and $\big(V^*DV\big)$ is hermitian and similar to $D$, so apply your interlacing statement to the leading $m\times m$ principle submatrix given by $S^*\big(V^*DV\big)S$ Jul 11 at 15:57

Alternative solution:

For a matrix $$A \in \mathbb{C}^{n\times n}$$ whose eigenvalues are all real, denote by $$\lambda_1(A) \ge \lambda_2(A) \ge \cdots \ge \lambda_n(A)$$ its eigenvalues.

We have \begin{align*} \det(W^\ast D W) &= \prod_{k=1}^m \lambda_k (W^\ast D W) \\ &= \prod_{k=1}^m \lambda_k (DWW^\ast) \tag{1}\\ &\le \prod_{k=1}^m \lambda_k (D) \cdot \prod_{k=1}^m \lambda_k(WW^\ast) \tag{2}\\ &= \prod_{k=1}^m d_k \cdot \prod_{k=1}^m \lambda_k(W^\ast W) \\ &\le \prod_{k=1}^m d_k \cdot \left(\frac{\sum_{k=1}^m \lambda_k(W^\ast W)}{m}\right)^m \tag{3}\\ &= \prod_{k=1}^m d_k \cdot \left(\frac{\mathrm{tr}(W^\ast W)}{m}\right)^m \\ &= \prod_{k=1}^m d_k. \tag{4} \end{align*} Explanations:
(1): All non-zero eigenvalues (counting algebraic multiplicity) of $$AB$$ and $$BA$$ are the same.
(2): Theorem H.1. in (Page 338, [1]).
(3): AM-GM inequality.
(4): $$\mathrm{tr}(W^\ast W) = m$$.

Also, when $$W$$ has orthonormal columns, we have $$\det(W^\ast D W) = \prod_{k=1}^m d_k$$.

Thus, the maximum of $$\det(W^\ast D W)$$ is $$\prod_{k=1}^m d_k$$.

Reference.

[1] A. W. Marshall, I. Olkin, and B. Arnold, “Inequalities: Theory of Majorization and Its Applications,” 2011.

• Nice. $W^•W$ is a Gram matrix of vectors with length $1.$ So its determinant is less or equal $1.$ Jul 7 at 5:57
• @RyszardSzwarc That's another way. Thanks. Jul 7 at 5:59
• @user8675309 Thanks for your suggestion regarding the use of more appropriate theorems. Jul 12 at 22:28