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Expert-verified Found in: Page 413 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider a singular value decomposition ${\mathbit{A}}{\mathbf{=}}{\mathbit{U}}{\mathbit{\Sigma }}{{\mathbit{V}}}^{{\mathbf{T}}}$of an ${\mathbit{n}}{\mathbf{×}}{\mathbit{m}}$matrix A with rank${\mathbit{A}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbit{m}}$. Let ${\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{\dots }}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{n}}}$be the columns of U . Without using the results of Chapter 5 , compute${\mathbit{A}}{\left({A}^{T}A\right)}^{\mathbf{-}\mathbf{1}}{{\mathbit{A}}}^{{\mathbf{T}}}{\stackrel{\mathbf{\to }}{\mathbf{u}}}_{{\mathbf{i}}}{\mathbf{.}}$ Explain your result in terms of Theorem 5.4.7.

$A{\left({A}^{T}A\right)}^{-1}{A}^{T}{\stackrel{˙}{u}}_{i}=\left\{\begin{array}{ll}{\stackrel{˙}{u}}_{i}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

See the step by step solution

## Step 1:To compute AATA-1ATu→i.

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

$\begin{array}{rcl}{A}^{T}{\stackrel{˙}{u}}_{i}& =& \left(U\Sigma {V}^{T}\right){\stackrel{˙}{u}}_{i}\\ & =& \left(V{\Sigma }^{T}{U}^{T}\right){\stackrel{˙}{u}}_{i}\\ & =& V{\Sigma }^{T}\left({U}^{T}{\stackrel{˙}{u}}_{i}\right)\\ & =& V{\Sigma }^{T}{e}_{i}\\ & & \end{array}$

where${\mathbf{e}}_{i}$ is the ${i}^{\text{th}}$ basis vector of ${\mathrm{ℝ}}^{n}.$

Since ${\Sigma }^{T}$ is an $n×m$ matrix

role="math" localid="1660679956475" ${\Sigma }^{T}{\mathbf{e}}_{i}=\left\{\begin{array}{ll}{\sigma }_{i}{\mathbf{e}}_{i}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

Hence,

role="math" localid="1660680127331" $V{\Sigma }^{T}{\mathbf{e}}_{i}=\left\{\begin{array}{ll}V{\sigma }_{i}{\mathbf{e}}_{i}\mathbf{}\mathbf{=}{\mathbf{\sigma }}_{\mathbf{i}}{\stackrel{\mathbf{˙}}{\mathbf{v}}}_{\mathbf{i}}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

Thus,

${A}^{T}{\stackrel{˙}{u}}_{i}=\left\{\begin{array}{ll}{\mathbf{\sigma }}_{\mathbf{i}}{\stackrel{\mathbf{˙}}{\mathbf{v}}}_{\mathbf{i}}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

## Step2: ToExplain the result in terms of Theorem 5.4.7.

Recall that ${\stackrel{˙}{v}}_{i}\text{'}s$ are eigenvectors of ${A}^{T}A$ with eigenvalues ${\sigma }_{i}^{2}.$

Hence, $\left({A}^{T}A\right){\stackrel{˙}{v}}_{i}={\sigma }_{i}^{2}{\stackrel{˙}{v}}_{i}.$This implies that ${\left({A}^{T}A\right)}^{-1}\left({\sigma }_{i}{\stackrel{˙}{v}}_{i}\right)=\frac{1}{{\sigma }_{i}}{\stackrel{˙}{v}}_{i}.$

This implies the following.

${\left({A}^{T}A\right)}^{-1}{A}^{T}{\stackrel{˙}{u}}_{i}=\left\{\begin{array}{ll}\frac{1}{{\sigma }_{i}}{\stackrel{˙}{v}}_{i}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

Now, use that

$A{\stackrel{˙}{v}}_{i}={\sigma }_{i}{\stackrel{˙}{u}}_{i}.$

Hence,

$A{\left({A}^{T}A\right)}^{-1}{A}^{T}{\stackrel{˙}{u}}_{i}=\left\{\begin{array}{ll}{\stackrel{˙}{u}}_{i}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$

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

## Step 3:Final proof

$A{\left({A}^{T}A\right)}^{-1}{A}^{T}{\stackrel{˙}{u}}_{i}=\left\{\begin{array}{ll}{\stackrel{˙}{u}}_{i}& \text{if}1⩽i⩽m\\ 0& \text{otherwise}\end{array}\right\$ ### Want to see more solutions like these? 