## 每週問題 January 23, 2017

Prove that if $A$ is a real symmetric positive definite then $\hbox{adj}A$ is also a symmetric positive definite matrix.

\begin{aligned} \hbox{adj}A&=\hbox{adj}(QDQ^T)=(\hbox{adj}Q^T)(\hbox{adj}D)(\hbox{adj}Q)\\ &=(\det Q^T)(Q^T)^{-1}(\hbox{adj}D)(\det Q)Q^{-1}\\ &=(\det Q)^{-1}Q(\hbox{adj}D)(\det Q)Q^T\\ &=Q(\hbox{adj}D)Q^T, \end{aligned}

$\hbox{adj}D=\begin{bmatrix} \prod_{i\neq 1}\lambda_i&&\\ &\ddots&\\ &&\prod_{i\neq n}\lambda_i \end{bmatrix}$