Q35E Consider a singular value decomp... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 8: Q35E (page 413)

Consider a singular value decomposition $A = U 危 V^{T}$ of an $n 脳 m$ matrix Awith $rank A = m$ . Let ${\overset{鈫�}{v}}_{1}, 鈥�, {\overset{鈫�}{v}}_{m}$ be the columns of Vand ${\overset{鈫�}{u}}_{1}, 鈥�, {\overset{鈫�}{u}}_{n}$ the columns of U. Without using the results of Chapter 5 , compute ${(A^{T} A)}^{- 1} A^{T} {\overset{鈫�}{u}}_{i}$ . Explain the result in terms of leastsquares approximations.

Short Answer

Expert verified

${(A^{T} A)}^{- 1} A^{T} {\overset{藱}{u}}_{i} = \{\begin{cases} \frac{1}{蟽_{i}} {\overset{藱}{v}}_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

Step by step solution

To compute ATA-1ATu→i

Let's first compute $A^{T} {\overset{藱}{u}}_{i}$

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

where $e_{i}$ is the $i^{th}$ basis vector of ${鈩�}^{n}$

Since $危^{T}$ is an $n 脳 m$ matrix,

$危^{T} e_{i} = \{\begin{cases} 蟽_{i} e_{i} = 蟽_{i} {\overset{藱}{v}}_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

Hence,

role="math" localid="1660678707172" $V 危^{T} e_{i} = \{\begin{cases} V 蟽_{i} e_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

Thus,

$A^{T} {\overset{藱}{u}}_{i} = \{\begin{cases} 蟽_{i} {\overset{藱}{v}}_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

To Explain the result in terms of least squares approximations.

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

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

This implies the following.

${(A^{T} A)}^{- 1} A^{T} {\overset{藱}{u}}_{i} = \{\begin{cases} \frac{1}{蟽_{i}} {\overset{藱}{v}}_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

The above result implies that the least square solution of $A \overset{藱}{x} = {\overset{藱}{u}}_{i} i s \frac{1}{蟽_{i}} {\overset{藱}{v}}_{i}$

Step 3:Final proof

${(A^{T} A)}^{- 1} A^{T} {\overset{藱}{u}}_{i} = \{\begin{cases} \frac{1}{蟽_{i}} {\overset{藱}{v}}_{i} & if 1 猢� i 猢� m \\ 0 & otherwise \end{cases}$

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91影视

Short Answer

Step by step solution

To compute ATA-1ATu→i

To Explain the result in terms of least squares approximations.

Step 3:Final proof

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