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.

`\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. $\endgroup$