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I want to check whether the given set is closed or bounded $$S = \{X\in \mathbb R^{2\times2}:X\succcurlyeq0\text{ and trace(}X) = 1\}$$

I tried it in the following manner:

Let the generic matrix be \begin{bmatrix} x_1 & x_2\\ x_3 & 1-x_1 \end{bmatrix}, Then we find the eigenvalues of the matrix to satisfy the P.S.D. condition. This gives us $\lambda^2-\lambda+(x_1-x_1^2-x_2x_3)=0$. Now, both the eigenvalues should be positive, and thus we get the condition $x_1-x_1^2-x_2x_3\geq0.25$

So now the set will be closed and bounded if the set S = $\{[x_1,x_2,x_3]\in\mathbb R : x_1-x_1^2-x_2x_3\geq0.25\}$ is closed and bounded.

But while calculating the critical point of the above function I found that it does not have any local/global minimum. The critical point is ($\frac{1}{2},0,0$) and results in the function value of 0.25

Can anyone give a hint about how to proceed further?

Is there any other method that is less calculation intensive?

Thanks!

amaths
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    What does $\succcurlyeq$ mean? Are P.S.D. matrices always symmetric? – kimchi lover Sep 05 '22 at 19:27
  • @kimchilover The symbol is for PSD. NO PSDs are not always symmetric. – amaths Sep 06 '22 at 03:17
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    @amaths: Then what does it mean to be PSD? – PhoemueX Sep 06 '22 at 13:27
  • @PhoemueX PSD is when uTxu is non-nrgative. PSD matrix need not always be symmetric. See this https://math.stackexchange.com/questions/1954167/do-positive-semidefinite-matrices-have-to-be-symmetric#:~:text=No%2C%20they%20don't%2C,(2022). – amaths Sep 07 '22 at 09:17

1 Answers1

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The condition for your eigenvalues to be positive is not $x_1-x_1^2-x_2x_3\geq0.25$ and anyway, the positivity of the eigenvalues is not a sufficient condition for the matrix to be PSD.

The eigenvalues of a matrix $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ are $\ge0$ iff their sum and product are, i.e. iff $a+d\ge0$ and $ad-bc\ge0$.

$X$ is PSD iff $a,d\ge0$ and $(b+c)^2\le4ad$, which is a strictly stronger constraint, even when $d=1-a$.

$$S=\left\{\left.\begin{pmatrix}a&b\\c&1-a\end{pmatrix}~\right|~0\le a\le1\text{ and }(b+c)^2\le4a(1-a)\right\}=\left\{\left.\begin{pmatrix}a&b\\c&1-a\end{pmatrix}~\right|~(b+c)^2\le4a(1-a)\right\}$$ is closed and unbounded:

  • closed as the inverse image of the closed set $\mathbb R_+$ by the continuous function $\mathbb R^3\to\mathbb R,\;(a,b,c)\mapsto4a(1-a)-(b+c)^2;$
  • unbounded because $\begin{pmatrix}0&n\\-n&0\end{pmatrix}\in S.$
Anne Bauval
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