Equivalence of minimizing trace and determinant over matrix quadratic form in multivariate regression

Consider the multivariate regression model $$Y = XB + E$$ where $$Y$$ is $$n \times p$$ and corresponds to the dependent variables, $$X$$ is $$n \times k$$ and corresponds to the independent variables, $$B$$ is $$k \times p$$ and corresponds to the coefficients that we wish to estimate. Lastly, let $$E \sim N_{n, p}(0, \Sigma, I_n)$$ correspond to the errors in the model. We can directly note that $$(Y-XB)^{T}(Y-XB) \succ 0$$.

I wonder if the following equivalence holds: \begin{align} & \operatorname*{argmin}_B \operatorname{trace}(Y-XB)^{T}(Y-XB) \\ = {} & \operatorname*{argmin}_B \det(Y-XB)^{T}(Y-XB). \end{align}

I have done some simulations using real data which seems to indicate that it is positive, but I'm very skeptical (since the left hand side is convex in $$B$$ and the right hand side isn't). Some aspects I have already considered are:

• Since we have strong convexity of the left hand side, we know that a minimizer exists. Denoting this by $$B^{*}$$ we have the sufficient condition $$(Y-XB^{*})^T(Y-XB^{*}) \preceq (Y-XB)^T(Y-XB)$$ which is also quite hard to prove (in the case that it is true).

• Using that $$\det(e^{(Y-XB)^{T}(Y-XB)}) = e^{\operatorname{trace}(Y-XB)^T(Y-XB)}$$ and considering the corresponding eigenvalues.

I would love any insights or counter-examples to the equivalence statement.

• When you write \text{det} A rather than \det A then you see $\text{det} A$ rather than $\det A.$ This doesn't just mean that \det provides horizontal spacing; rather the spacing depends on the context, so that \det (A) gives you $\det(A),$ with a smaller amount of space to the right. And this also applies to the left side. The same applies to using \text{trace} rather than \operatorname{trace}. And in \operatorname*{argmin}, the asterisk affects the formatting of the subscript when in a "displayed" (as opposed to "inline") context. Commented Apr 22 at 17:51
• \begin{align} & \text{det} A \\ \text{versus } \qquad & {\det A} \end{align} Commented Apr 22 at 17:52

$$\newcommand\tr{\operatorname{tr}}$$Let $$Q(B):=(Y-XB)^T(Y-XB).$$ Since the column spaces of the matrices $$X^TX$$ and $$X^T$$ are the same, there is a matrix $$B_*$$ such that $$X^TXB_*=X^T Y.$$ For each $$z\in\Bbb R^{p\times1}$$, by (say) differentiating the convex function $$B\mapsto z^T Q(B)z$$ in $$B$$, we see that $$B_*$$ is a minimizer of this function. So, $$B_*$$ is a minimizer of $$Q(B)$$ in $$B$$ wrt to the Loewner ordering. So, $$B_*$$ is a minimizer of both $$\tr Q(B)$$ and $$\det Q(B)$$ in $$B$$.
Generically, $$\tr Q(B)$$ is strictly convex in $$B$$, and then $$B_*$$ is a unique minimizer of $$\tr Q(B)$$. However, a minimizer of $$\det Q(B)$$ is not unique.
For instance, suppose that $$X=\left( \begin{array}{cc} 1 & 3 \\ 1 & 1 \\ \end{array} \right), \quad Y=\left( \begin{array}{cc} 3 & 0 \\ 2 & 1 \\ \end{array} \right).$$ Then $$B_*=\frac12\,\left( \begin{array}{cc} 3 & 3 \\ 1 & -1 \\ \end{array} \right)$$ is the unique minimizer of $$\tr Q(B)$$. However, $$\left( \begin{array}{cc} 0 & 0 \\ 0 & -1 \\ \end{array} \right)$$ is a minimizer of $$\det Q(B)$$.
• Thank you. Since the Loewner order is only a partial ordering, can we actually make the conclusion that $B_{*}$ is a minimizer (wrt. the ordering) over all $k \times p$ matrices $B$? Commented Apr 23 at 7:13
• @respectableuser1 : It is shown here that $Q(B_*)\lesssimeq Q(B)$ for all $B$, where $\lesssimeq$ is the Loewner order. That $\lesssimeq$ is a partial order does not diminish this fact. Somewhat similarly, $0\lesssimeq B^T B$ for all $B$, even though $\lesssimeq$ is only a partial order. Do you see it now? Commented Apr 24 at 1:24