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  • [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)$ a$(2). Since $A$ is positive, and thus is Hermitian, $exists$ unitary $Q$, $st$ $$ex A=Qdiag(lm_1,cdots,lm_n)Q^*,quad lm_igeq 0. eex$$ Take $$ex B=diagsex{sqrt{lm_1},cdots,sqrt{lm_n}}Q, eex$$ then $A=B^*B$.

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

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

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

    (1)$ a$(3). Just use the QR decomposition.

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  • 原文地址:https://www.cnblogs.com/zhangzujin/p/4105333.html
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