by svart
Last Updated May 15, 2018 20:20 PM

Given a $n\times n$ P.D. matrix $A$ and a diagonal matrix $D = \mathrm{diag}(d_1,\ldots,d_n)$, with $d_i > 0$, $i=1\ldots,n$, I want to prove that the product $AD$ is also positive definite.

Since $A$ is positive definite, all its leading principal minors will be positive.

Let with $|\cdot|_k$ denote the $k$-th leading principal minor and let with $[\cdot]_{R,C}$, where $R, C$ are $k$-subsets of $\{1,\ldots,n\}$, denote the $k\times k$ minor that corresponds to the rows with index in $R$ and the columns with index in $C$.

From the formula in Wikipedia, we get $$ |AD|_k = [AD]_{I,I} \sum_K [A]_{I,K} [D]_{K,I}, $$ where $I = \{1, \ldots, k\}$ and the sum extends over all subsets $K$ of $\{1,...,n\}$ with $k$ elements.

However, since $D$ is diagonal, every minor with $K \neq I$ will be zero. Thus, $$ |AD|_k = [A]_{I,I} [D]_{I,I} = |A|_k |D|_k, $$ which will be positive for all $k$, and as such $AD$ is positive definite.

Is my reasoning correct?

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