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.


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.

  • $\begingroup$ 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. $\endgroup$ Jun 10, 2016 at 7:25
  • $\begingroup$ @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. $\endgroup$
    – user251057
    Jun 10, 2016 at 7:30
  • $\begingroup$ A matrix and its transpose have the same eigenvalues. But you didn't answer my question. $\endgroup$ Jun 10, 2016 at 9:12
  • $\begingroup$ Try showing that $B'B$ is positive definite, it's easier. ;-) $\endgroup$
    – egreg
    Jun 10, 2016 at 10:40

1 Answer 1



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.


$x^\top B^\top Bx=(Bx)^\top Bx=||Bx||^2\ge0$

  • $\begingroup$ For the first question A cannot have negative eigen values since A is given as positive definite. $\endgroup$
    – user251057
    Jun 10, 2016 at 8:04

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