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### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question: 13. Exercises 12–14 concern an $$m \times n$$ matrix $$A$$ with a reduced singular value decomposition, $$A = {U_r}D{V_r}^T$$, and the pseudoinverse $${A^ + } = {U_r}{D^{ - 1}}{V_r}^T$$.Suppose the equation $$A{\rm{x}} = {\rm{b}}$$ is consistent, and let $${{\rm{x}}^ + } = {A^ + }{\rm{b}}$$. By Exercise 23 in Section 6.3, there is exactly one vector $${\rm{p}}$$ in Row $$A$$ such that $$A{\rm{p}} = {\rm{b}}$$. The following steps prove that $${{\rm{x}}^ + } = {\rm{p}}$$ and $${{\rm{x}}^ + }$$is the minimum length solution of $$A{\rm{x}} = {\rm{b}}$$.Show that $${{\rm{x}}^ + }$$ is in Row $$A$$. (Hint: Write $${\rm{b}}$$ as $$A{\rm{x}}$$ for some $${\rm{x}}$$, and use Exercise 12.) Show that $${{\rm{x}}^ + }$$ is a solution of $$A{\rm{x}} = {\rm{b}}$$.Show that if $${\rm{u}}$$ is any solution of $$A{\rm{x}} = {\rm{b}}$$, then $$\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|$$, with equality only if $${\rm{u}} = {{\rm{x}}^ + }$$.

1. It is shown that $${{\rm{x}}^ + }$$ is in Row $$A$$.
2. It is shown that $${{\rm{x}}^ + }$$ is the solution of $$A{\rm{x}} = {\rm{b}}$$.
3. It is shown that $$\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|$$, and equality holds if $${\bf{u}} = {{\rm{x}}^ + }$$.
See the step by step solution

## Step 1: Simplify for vector $${{\rm{x}}^ + }$$

It is given that the system $$A{\rm{x}} = {\rm{b}}$$ is consistent and $${{\rm{x}}^ + } = {A^ + }{\rm{b}}$$, then the system $$A{\rm{x}} = {\rm{b}}$$ can be simplified as follows:

$$\begin{array}{c}{{\rm{x}}^ + } = {A^ + }{\rm{b}}\\ = {A^ + }A{\rm{x}}\end{array}$$

As $${{\rm{x}}^ + } = {A^ + }A{\rm{x}}$$, so$${{\rm{x}}^ + }$$ is the orthogonal projection of $${\rm{x}}$$ onto Row $$A$$.

## Step 2: Simplify for product $$A{{\rm{x}}^ + }$$

Substitute $${{\rm{x}}^ + } = {A^ + }A{\rm{x}}$$, into $$A{{\rm{x}}^ + }$$, and simplify using $$A{\rm{x}} = {\rm{b}}$$, as follows:

$$\begin{array}{c}A{{\rm{x}}^ + } = A\left( {{A^ + }A{\rm{x}}} \right)\\ = A{A^ + }A{\rm{x}}\\ = A{\rm{x}}\\ = {\rm{b}}\end{array}$$

Thus, it is shown that $${{\rm{x}}^ + }$$ is the solution of $$A{\rm{x}} = {\rm{b}}$$.

## Step 3: Prove $$\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|$$

Suppose the system $$A{\rm{x}} = {\rm{b}}$$ is satisfied by another basis $${\rm{u}}$$, such that, $$A{\rm{u}} = {\rm{b}}$$. It is also known that $${{\rm{x}}^ + }$$is the orthogonal projection of $${\rm{x}}$$ onto Row $$A$$.

Apply the Pythagorean theorem on $${\left\| {\rm{u}} \right\|^2}$$, as follows:

$$\begin{array}{c}{\left\| {\rm{u}} \right\|^2} = {\left\| {{{\rm{x}}^ + }} \right\|^2} + {\left\| {{\rm{u}} - {{\rm{x}}^ + }} \right\|^2}\\ \ge {\left\| {{{\rm{x}}^ + }} \right\|^2}\end{array}$$

So $$\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|$$, and equality holds if $${\bf{u}} = {{\rm{x}}^ + }$$.