Short Answer

Consider a QR factorization
M=QR Show that $R = Q^{T} M$ .

$R = Q^{T} M$

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Step by Step Solution

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TABLE OF CONTENTS

Step 1: QR factorization.

Consider an n × m matrix M with linearly independent column ${\overset{⊥}{v}}_{1}, . . . ., {\overset{⊥}{v}}_{m}$ .Then there exists an n × m matrix Q whose columns ${\overset{1}{u}}_{1}, . . . ., {\overset{1}{u}}_{m}$ are orthonormal and an upper triangular matrix R with positive diagonal entries such that.

M=QR

To prove that $R = Q^{T} M$

Since, the column of the matrix Q is orthonormal and therefore it gives,

$Q Q^{T} = l = Q^{T} Q$

Thus, it is written as,

$M = Q R \Rightarrow Q^{T} M = Q^{Q} R \Rightarrow Q^{T} M = I R \Rightarrow R = Q^{T} M$

Hence, $R = Q^{T} M$ proved.

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Linear Algebra With Applications

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Short Answer

Consider a QR factorization
M=QR Show that $R = Q^{T} M$ .

Step by Step Solution

Step 1: QR factorization.

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Linear Algebra With Applications

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Consider a QR factorization
M=QR Show that $R = Q^{T} M$ .