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Exercise 1.10.5

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When $D = Λ^{- \frac{1}{2}}$ , we have $D^{T} ΛD = Λ^{- \frac{1}{2}} Λ Λ^{- \frac{1}{2}} = I$ .

By definition, $Q_{2}$ is the eigenvector matrix of the matrix $D^{T} Q^{T} MQD$ , this matrix is symmetric, so we have $D^{T} Q^{T} MQD = Q_{2} Λ_{2} Q_{2}^{T}$ , and $Q_{2}^{T} D^{T} Q^{T} MQD Q_{2} = Λ_{2}$ .

We have $Z^{T} SZ = Q_{2}^{T} D^{T} Q^{T} SQD Q_{2} = Q_{2}^{T} I Q_{2} = I$ , and $Z^{T} MZ = Q_{2}^{T} D^{T} Q^{T} MQD Q_{2} = Λ_{2}$ . So $Z$ can diagonalize both congruences $Z^{T} SZ$ and $Z^{T} MZ$ .

niuers

2020-03-20 00:00

Exercise 1.10.5

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