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.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 22 at 17:51
  • $\begingroup$ $$ \begin{align} & \text{det} A \\ \text{versus } \qquad & {\det A} \end{align}$$ $\endgroup$ Commented Apr 22 at 17:52

1 Answer 1


$\newcommand\tr{\operatorname{tr}}$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 $\det Q(B)$.

  • $\begingroup$ 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 $B$? $\endgroup$ Commented Apr 23 at 7:13
  • $\begingroup$ @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? $\endgroup$ Commented Apr 24 at 1:24
  • $\begingroup$ Yes I see it. Thank you very much for your time. $\endgroup$ Commented Apr 24 at 7:03

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