[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.2

Show that the following statements are equivalent:

(1). $A$ is positive.

(2). $A=B^*B$ for some $B$.

(3). $A=T^*T$ for some upper triangular $T$.

(4). $A=T^*T$ for some upper triangular $T$ with nonnegative diagonal entries. If $A$ is positive definite, then the factorization in (4) is unique. This is called the Cholesky decomposition of $A$.

Solution.  (1)$\ra$(2). Since $A$ is positive, and thus is Hermitian, $\exists$ unitary $Q$, $\st$ $$\bex A=Q\diag(\lm_1,\cdots,\lm_n)Q^*,\quad \lm_i\geq 0. \eex$$ Take $$\bex B=\diag\sex{\sqrt{\lm_1},\cdots,\sqrt{\lm_n}}Q, \eex$$ then $A=B^*B$.

(2)$\ra$(4). By QR decomposition, $\exists$ orthogonal $Q$, upper triangular $R$ with diagonals $\geq0$, $\st B=QR$. Thus $$\bex A=B^*B=R^*Q^*QR=R^*R. \eex$$

(4)$\ra$(1). First, $A$ is Hermitian. Second, $$\bex x^*Ax=x^*T^*Tx=\sen{Tx}^2\geq 0,\quad \forall\ x. \eex$$

(3)$\ra$(1). Just do as that in (4)$\ra$(1).

(1)$\ra$(3). Just use the QR decomposition.