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Q13SE

Expert-verifiedFound in: Page 395

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 \({\rm{u}} = {{\rm{x}}^ + }\).**

- It is shown that \({{\rm{x}}^ + }\) is in Row \(A\).
- It is shown that \({{\rm{x}}^ + }\) is the solution of \(A{\rm{x}} = {\rm{b}}\).
- It is shown that \(\left\| {{{\rm{x}}^ + }} \right\| < \left\| {\rm{u}} \right\|\), and equality holds if \({\bf{u}} = {{\rm{x}}^ + }\).

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 \(A\).

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 \(A{\rm{x}} = {\rm{b}}\).

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 \({\bf{u}} = {{\rm{x}}^ + }\).

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