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Let $X$ be a square matrix. Is $\operatorname{trace}(X)$ a smooth function of $X$? Why?

What if $X$ positive-semi-definite?

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The trace and determinant functions are polynomials in the entries of the matrix and therefore are smooth. – Bill Cook Feb 15 '12 at 2:49
What's the domain you have in mind? The set of all square matrices? If that's the case, then $trace$ is a linear functional in a finite-dimensional vector space, so it is smooth. – student Feb 15 '12 at 2:49
up vote 1 down vote accepted

Any polynomial $\mathbb{R}[x_1,\dots,x_N]$ is a smooth function (it will have continuous partials of all orders).

The trace function is given by $\mathrm{Tr}(X) = \mathrm{Tr}\left(\begin{bmatrix} x_{11} & \cdots & x_{1n} \\ \vdots & & \vdots \\ x_{n1} & \cdots & x_{nn} \end{bmatrix} \right) = x_{11}+x_{22}+\cdots+x_{nn}$. This is a (linear) polynomial in the $x_{ij}$'s -- that is -- $\mathrm{Tr}(X) \in \mathbb{R}[x_{11},\dots,x_{1n},x_{21},\dots,x_{2n},\dots,x_{n1},\dots,x_{nn}]$ (call this $\mathbb{R}[x_{ij}]_{1 \leq i,j \leq n}$ for short. Thus the trace map to smooth because it's a polynomial.

The same can be said of the determinant. Denote the symmetric group as $S_n$ and let $(-1)^\sigma$ be the sign of $\sigma \in S_n$. Then $\mathrm{det}(X) = \sum\limits_{\sigma \in S_n} (-1)^\sigma x_{1\sigma(1)}\cdots x_{n\sigma(n)}$ which is a (homogeneous degree $n$) polynomial. Thus the determinant is smooth as well.

Restricting these maps to positive semi-definite matrices will still leave us with smooth maps.

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