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Q46E

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Found in: Page 234

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Consider a QR factorizationM=QR Show that ${\mathbit{R}}{\mathbf{=}}{{\mathbit{Q}}}^{{\mathbf{T}}}{\mathbit{M}}$.

$R={Q}^{T}M$

See the step by step solution

## Step 1: QR factorization.

Consider an n × m matrix M with linearly independent column ${\stackrel{\mathbf{\perp }}{\mathbf{v}}}_{{1}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{\perp }}{\mathbf{v}}}_{{\mathbf{m}}}$.Then there exists an n × m matrix Q whose columns ${\stackrel{\mathbf{1}}{\mathbf{u}}}_{1}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{1}}{\mathbf{u}}}_{{\mathbf{m}}}$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,

$Q{Q}^{T}=l\phantom{\rule{0ex}{0ex}}={Q}^{T}Q$

Thus, it is written as,

$M=QR\phantom{\rule{0ex}{0ex}}⇒{Q}^{T}M={Q}^{Q}R\phantom{\rule{0ex}{0ex}}⇒{Q}^{T}M=IR\phantom{\rule{0ex}{0ex}}⇒R={Q}^{T}M$

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