which of the following are true regarding the positive deﬁnite ness of matrices Pick out the true statements:
(a) Let $A$ be a hermitian $N × N$ positive deﬁnite matrix. Then, there exists
a hermitian positive deﬁnite $N × N$ matrix $B$ such that $B^2 = A$.
(b) Let $B$ be a nonsingular $N × N$ matrix with real entries. Let $B′$ be its
transpose. Then $B′B$ is a symmetric and positive deﬁnite matrix.

totally stuck.How to solve this problem.
 A: They are both true.
For (a), since $A$ is Hermitian, there exists a unitary matrix $U$ diagonalizing $A$, thus:
$U^\dagger A U = \Lambda$, where $\Lambda$ is a diagonal matrix with the eigenvalues $\lambda_1, \lambda_2, . . . , \lambda_N$ on the main diagonal and zeroes everywhere else.  Since $A^\dagger = A$, the $\lambda_i$ are all real.  Since $A$ is positive definite, $\lambda_i > 0$ for all $i, 1 \le i \le N$, since we may write $Ax = \lambda_i x$ for a unit eigenvector $x$ corresponding to $\lambda_i$, and thus have $\lambda_i = \lambda_i \langle x, x \rangle = \langle x, \lambda_i x \rangle = \langle x, Ax \rangle  > 0$.  Now let $K$ be the diagonal matrix $\text{diag}(\sqrt \lambda_1, \sqrt \lambda_2, . . ., \sqrt \lambda_N)$; then clearly $K^2 = \Lambda$, whence
$A = U \Lambda U^\dagger = U K^2 U^\dagger = U K U^\dagger U K U^\dagger = (U K U^\dagger)^2 = B^2, \tag{1}$
where $B = U K U^\dagger$; then $B^\dagger = U K^\dagger U^\dagger = U K U^\dagger = B$, so $B$ is Hermitian.  Finally, $B$ is positive definite since its eigenvalues are the $\sqrt \lambda_i > 0$.
Thus we assign statement (a) the truth value TRUE.
For (b), notice that $(B'B)' = B'B'' = B'B$, so $B'B$ is symmetric.  For any vector $x \ne 0$, we have $\langle x, B'Bx \rangle = \langle Bx, Bx \rangle > 0$ since $B$ is nonsingular; this shows $B'B$ is positive definite.
Thus we assign statement (b) the truth value TRUE.
Hope this helps.  Have a Happy New Year,
and as always,
Fiat Lux!!!
