# How can i show this Function is Coercive?

Let $M_n$ denote the set of real matrices with real entries and order $n$. Let $D=\{X\in M_n:\ \det(X)>0 \}$, where $\det$ stands for determinant. Define $F:D\rightarrow\mathbb{R}$ by $$F(X)=\operatorname{tr}(X)-\lambda\ln(\det(X))$$

where $\lambda>0$ and $\operatorname{tr}(X)$ is the trace of $X$.

I want to show that $F$ is coercive, i.e. if $X_k\rightarrow \partial D$ or $\|X_k\|\rightarrow\infty$ then $\limsup F(X_k)=+\infty$.

Thanks

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Which norm are you using? If it is the largest eigenvalue then bound the trace below by the largest eigenvalue and the determinant above by that eigenvalue to the $n^{th}$ power. If the limit goes to zero, the trace term goes to zero and the log goes to negative infinity. – Chris Janjigian Nov 27 '12 at 13:26
You can take the norm thats makes it easy. – Tomás Nov 27 '12 at 13:31

The claim is false as stated. Let $n = 2$, consider $X_n = \begin{pmatrix} -n & 0 \\ 0 & -n \end{pmatrix}$. Since $\det X_n = n^2 > 0$ we have that $X_n \in D$. By positive homogeneity of norms $\|X_n\| = n \|X_1\| \nearrow \infty$. But $\ln \det X_n = 2 \ln n$, while $\operatorname{tr} X_n = -2n$, so $F(X_n) = - 2n - 2 \ln n \searrow -\infty$.

Edit [A slightly different counterexample]:

Again take $n = 2$. Take $X_n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$. We have that the trace of $X_n$ is always 2, while the determinant is always 1. So $F(X_n) = 2$ for all $n$. Both the Frobenius norm and the max norm on $X_n$ satisfy $\|X_n \| \geq n$ and this gives another family of counterexamples.

Furthermore, this example gives also an illustration of why you need to be precise about $X_k \to \partial D$. Observe that the matrix $Y_n = \begin{pmatrix} 1 & n \\ 1/n & 1\end{pmatrix}$ has vanishing determinant, and hence belongs in $\partial D$. In any norm $$\|X_n - Y_n\| = \left\| \begin{pmatrix} 0 & 0 \\ 1/n & 0\end{pmatrix} \right\| \searrow 0$$ so we in fact have not only do $X_n$ blow up in norm, it also asymptotically approaches $\partial D$!

(On the other hand, the statement that if $X_n \to X_0\in \partial D$ we have $F(X_n) \nearrow \infty$ is true by virtue of continuity of the trace function on $M_n$.)

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With minor modifications the same counterexample works for any $n$. To make it work you need to restrict to a domain $D$ such that $\| X_k \|$ provides a lower bound for the trace. – Willie Wong Nov 27 '12 at 13:44
Thank you Willie – Tomás Nov 27 '12 at 13:53

Both the trace and determinant of a matrix are invariant under conjugation, and the diagonalizable matrices with positive determinant are dense in $D$, so it suffices to show the result for that dense subset. Thus we can consider the question be to:

Let $F(a_1,\cdots, a_n) = a_1+\cdots +a_n -\lambda \log(a_1 \cdots a_n)$ be a function of $n$ real variables where $a_1\cdots a_n>0.$ If some $a_i\to 0$ or $\infty$ then $F(a_1, \cdots, a_n) \to \infty.$

Let us write $F= \sum (a_i -\lambda \log a_i).$ If $a_k\to 0$ then $a_k -\lambda \log a_k$ tends to $+\infty$ since the log term blows up. If $a_k\to \infty$, the $a_k$ term blows up quicker than the log term decays, so again, $+\infty.$ And we are done.

As Willie Wong shows, the problem is false as stated. Where I went wrong above in my translation into a new problem is that where I wrote "If some $a_i\to 0$ or $\infty$" I should have wrote "If some $a_i\to 0$ or $\pm\infty$" - the case where $a_n \to -\infty$ is precisely where the above argument could fail.

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Actually, $a_n$ can remain bounded even as the norm blows up. Matrix norms give upper bounds to (absolute values of) eigenvalues but in general not corresponding lower bounds. – Willie Wong Nov 27 '12 at 13:50