Problem in solving a question of positive definite matrix.

The question is : Pick out the true statements :

(a)Let $A$ be a hermitian $N \times N$ positive definite matrix.Then, there exists a hermitian positive definite $N \times N$ matrix B such that $B^2 = A$.

(b)Let B be a non-singular $N \times N$ matrix with real entries.Let $B'$ be its transpose.Then $B'B$ is a symmetric and positive definite matrix.

Workdone:

For the first case I cannot make any conclusion.But for the second case though B'B is symmetric it need not be positive definite.Is it correct at all? Please check! Thank you in advance.

• Do you have an example where $B'B$ is not positive definite? If you do, then you don't need any help with that one. Jun 10, 2016 at 7:25
• @Gerry Myerson if eigen values of A are known then how can I find out the eigen values of the transpose of A say A'.Please make me aware about this fact.
– user251057
Jun 10, 2016 at 7:30
• A matrix and its transpose have the same eigenvalues. But you didn't answer my question. Jun 10, 2016 at 9:12
• Try showing that $B'B$ is positive definite, it's easier. ;-) Jun 10, 2016 at 10:40

As $A$ is Hermitian, it's unitarily diagonalizable i.e. $\exists U,D:U^*U=I$ and $D$ is diagonal and $A=UDU^{*}$. As $A$ is positive-definite, diagonal entries of $D$ are positive and we're able to define $E=D^{1\over2}$ as $E_{ij}=+\sqrt{D_{ij}}$. Clearly $E^2=D$. Let $B=UEU^{*}$. Then we have $$B^2=UEU^{*}UEU^{*}=UEIEU^{*}=UEEU^{*}=UDU^{*}=A\\ B^{*}=(U^{*})^{*}E^{*}U^{*}=UEU^{*}=B$$ So $B$ is Hermitian as $E$ is real and $B$ is positive definite as it's eigenvalues i.e. the diagonal entries of $E$ are positive.
$x^\top B^\top Bx=(Bx)^\top Bx=||Bx||^2\ge0$