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Exercise 3.3.7

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Since $A = QΛ Q^{- 1}$ , so we have $Q^{- 1} = {\bar{Q}}^{T}$ .

$Ā^{T} A = {(\bar{Q} \bar{Λ} Q^{T})}^{T} QΛ {\bar{Q}}^{T} = Q {\bar{Λ}}^{T} {\bar{Q}}^{T} QΛ {\bar{Q}}^{T} = Q \bar{Λ} Λ {\bar{Q}}^{T} = Q | Λ |^{2} {\bar{Q}}^{T}$

$A Ā^{T} = QΛ {\bar{Q}}^{T} {(\bar{Q} \bar{Λ} Q^{T})}^{T} = QΛ {\bar{Q}}^{T} Q {\bar{Λ}}^{T} {\bar{Q}}^{T} = QΛ \bar{Λ} {\bar{Q}}^{T} = Q | Λ |^{2} {\bar{Q}}^{T}$

So ${¯A}^TA =A{¯A}^T $

niuers

2020-03-20 00:00

Exercise 3.3.7

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