TITLE: Finding the number of symmetric, positive definite $10 \times 10$ matrices having...
QUESTION [5 upvotes]: I was looking at old exam papers and I was stuck with the following problem:

What is the number of symmetric, positive definite $10 \times 10$ matrices having trace equal to $10$ and determinant equal to $1$ ? The options are:

$0$

$1$

greater than $1$ but finite

Infinite.



After getting positive feedback,I am posting my attempt as a separate answer.

REPLY [6 votes]: If $A$ is positive definite, then its eigenvalues $\lambda_i$ are real and positive. We also know  that for any matrix, $\sum \lambda_i = \operatorname{trace}(A)$ and $\prod \lambda_i= |A|$. In this case, that means that we are restricted to $\sum \lambda_i =10$ and $\prod \lambda_i =1$. Now, applying $$A.M. \ge G.M. \implies \frac{\sum \lambda_i }{10} \ge \sqrt[10] {\prod \lambda_i} \implies 1 \ge 1,$$ where equality sign occurs only when $\lambda_1= \dots =\lambda_{10}$. Since in this case ,$1 \ngtr 1$, so only possibility is  when $1=1 \implies \lambda_1= \dots =\lambda_{10}.$
Hence only one such  symmetric, positive definite matrix (having all eigenvalues equal) is possible. So, option 2 is the right choice. 
As Sami Ben Romdhane mentioned: "The only matrix similar to the identity matrix is itself."