Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Consider a singular value decomposition $A = U Σ V^{T}$ of an $n \times m$ matrix A with rank $A = m$ . Let ${\vec{u}}_{1}, \dots, {\vec{u}}_{n}$ be the columns of U . Without using the results of Chapter 5 , compute $A {(A^{T} A)}^{- 1} A^{T} {\vec{u}}_{i} .$ Explain your result in terms of Theorem 5.4.7.

$A {(A^{T} A)}^{- 1} A^{T} {\dot{u}}_{i} = \{\begin{cases} {\dot{u}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:To compute AATA-1ATu→i.

Let's first compute $A^{T} {\dot{u}}_{i} .$

$\begin{array}{rcl} A^{T} {\dot{u}}_{i} & = & (U Σ V^{T}) {\dot{u}}_{i} \\ = & (V Σ^{T} U^{T}) {\dot{u}}_{i} \\ = & V Σ^{T} (U^{T} {\dot{u}}_{i}) \\ = & V Σ^{T} e_{i} \end{array}$

where $e_{i}$ is the $i^{th}$ basis vector of $ℝ^{n} .$

Since $Σ^{T}$ is an $n \times m$ matrix

role="math" localid="1660679956475" $Σ^{T} e_{i} = \{\begin{cases} σ_{i} e_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

Hence,

role="math" localid="1660680127331" $V Σ^{T} e_{i} = \{\begin{cases} V σ_{i} e_{i} = σ_{i} {\dot{v}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

Thus,

$A^{T} {\dot{u}}_{i} = \{\begin{cases} σ_{i} {\dot{v}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

Step2: ToExplain the result in terms of Theorem 5.4.7.

Recall that ${\dot{v}}_{i}' s$ are eigenvectors of $A^{T} A$ with eigenvalues $σ_{i}^{2} .$

Hence, $(A^{T} A) {\dot{v}}_{i} = σ_{i}^{2} {\dot{v}}_{i} .$ This implies that ${(A^{T} A)}^{- 1} (σ_{i} {\dot{v}}_{i}) = \frac{1}{σ_{i}} {\dot{v}}_{i} .$

This implies the following.

${(A^{T} A)}^{- 1} A^{T} {\dot{u}}_{i} = \{\begin{cases} \frac{1}{σ_{i}} {\dot{v}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

Now, use that

$A {\dot{v}}_{i} = σ_{i} {\dot{u}}_{i} .$

Hence,

$A {(A^{T} A)}^{- 1} A^{T} {\dot{u}}_{i} = \{\begin{cases} {\dot{u}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

If we are considering the sub space of, say W, which is spanned by $\{{\dot{u}}_{1}, {\dot{u}}_{2}, \dots, {\dot{u}}_{m}\}$ , then the above result implies that the projection ${\dot{u}}_{i}$ onto W is ${\dot{u}}_{i}$ if $1 ⩽ i ⩽ m$ and is 0 if $m + 1 ⩽ i ⩽ n .$

Step 3:Final proof

$A {(A^{T} A)}^{- 1} A^{T} {\dot{u}}_{i} = \{\begin{cases} {\dot{u}}_{i} & if 1 ⩽ i ⩽ m \\ 0 & otherwise \end{cases}$

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1:To compute AATA-1ATu→i.

Step2: ToExplain the result in terms of Theorem 5.4.7.

Step 3:Final proof

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Linear Algebra With Applications

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Short Answer

Consider a singular value decomposition A=UΣVTof an n×mmatrix A with rankA = m. Let u→1,…,u→nbe the columns of U . Without using the results of Chapter 5 , computeA(ATA)-1ATu→i. Explain your result in terms of Theorem 5.4.7.

Step by Step Solution

Step 1:To compute AATA-1ATu→i.

Step2: ToExplain the result in terms of Theorem 5.4.7.

Step 3:Final proof

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