The essential question to ask is what is \((H+\mu I)^{-1}\). We only need to show that \((Q\Lambda Q^{\top}+\mu I)^{-1}=Q(\Lambda+\mu I)^{-1}Q^{-1}\), which is equivalent to show \((Q\Lambda Q^{\top}+\mu I)Q(\Lambda+\mu I)^{-1}Q^{-1}=I\), then it is straightforward to get the conclusion. Use the fact that Q is orthonormal \((Q\Lambda Q^{\top}+\mu I) Q(\Lambda+\mu I)^{-1}Q^{-1}=(Q\Lambda +\mu Q) (\Lambda+\mu I)^{-1}Q^{-1}=Q(\Lambda +\mu I) (\Lambda+\mu I)^{-1}Q^{-1}=QQ^{-1}=I\).

## Lab 9 Problem 3

Hello,

in the solutions it says that inserting H = QLQ^T into w = (H + muI)^-1 H w* leads to the above solution,

but I dont see how

w = (QLQ^T - mu I )^-1 Q L Q^T w* reduces to that.

Thanks!

The essential question to ask is what is \((H+\mu I)^{-1}\). We only need to show that \((Q\Lambda Q^{\top}+\mu I)^{-1}=Q(\Lambda+\mu I)^{-1}Q^{-1}\), which is equivalent to show \((Q\Lambda Q^{\top}+\mu I)Q(\Lambda+\mu I)^{-1}Q^{-1}=I\), then it is straightforward to get the conclusion. Use the fact that Q is orthonormal

\((Q\Lambda Q^{\top}+\mu I) Q(\Lambda+\mu I)^{-1}Q^{-1}=(Q\Lambda +\mu Q) (\Lambda+\mu I)^{-1}Q^{-1}=Q(\Lambda +\mu I) (\Lambda+\mu I)^{-1}Q^{-1}=QQ^{-1}=I\).

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