The Number of symmetric,Positive Definite, $8\times 8$ matrices having trace$=8$ and determinant$=1$ is
$0$
$1$.
$>1$ but finite.
$\infty$
I am not able to do this one.
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If $A$ is pos. def. then its eigenvalues $\lambda_i$ are real and positive. Besides, we know (don't we?) that, for any matrix, $\sum \lambda_i = tr(A)$ and $\prod \lambda_i= |A|$. In our case, that means that we are restricted to $\sum \lambda_i =8$ and $\prod \lambda_i =1$... (can you go on from here? hint: arithmetic-geometric means and their properties) |
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