Question

Question: Let\(y = \left( {\begin{aligned}{}7\\9\end{aligned}} \right)\),\({{\bf{u}}_1} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\\{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\), and\(W = {\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\).

1. Let \(U\)be the\(2 \times 1\)matrix whose only column is\({{\bf{u}}_1}\). Compute\({U^T}U\)and\(U{U^T}\).
2. Compute\({\bf{Pro}}{{\bf{j}}_W}y\)and\(\left( {{U^T}U} \right){\bf{y}}\).

Short answer

1. 1. The matrices are\({U^T}U = \left( 1 \right)\),\(U{U^T} = \left( {\begin{aligned}{}{{1 \mathord{\left/{\vphantom {1 {10}}} \right.} {10}}}&{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}\\{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}&{{9 \mathord{\left/{\vphantom {9 {10}}} \right.} {10}}}\end{aligned}} \right)\).
   2. \({\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}} = \left( {\begin{aligned}{}{ - 2}\\6\end{aligned}} \right)\).

Step-by-step solution

Form \(U = \left( {{{\bf{u}}_1}} \right)\)

(a)

Given vector is\({{\bf{u}}_1} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\\{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\).

Find\(U = \left( {{{\bf{u}}_1}} \right)\).

\(U = \left( {\begin{aligned}{}{{1 \mathord{\left/{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\\{ - {3 \mathord{\left/{\vphantom {3 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\)

Determine \({U^T}U\) and \(U{U^T}\)

First, find the transpose of\(U\), that is\({U^T}\).

\({U^T} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}&{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\)

Now, compute\({U^T}U\)and\(U{U^T}\).

\(\begin{aligned}{c}{U^T}U = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}&{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\\{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.} {\sqrt {10} }}}\end{aligned}} \right)\\ = \left( {\frac{1}{{10}} + \frac{9}{{10}}} \right)\\ = \left( 1 \right)\end{aligned}\)

And,

\(\begin{aligned}{c}U{U^T} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.

\kern-\nulldelimiterspace} {\sqrt {10} }}}\\{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.

\kern-\nulldelimiterspace} {\sqrt {10} }}}\end{aligned}} \right)\left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {\sqrt {10} }}} \right.

\kern-\nulldelimiterspace} {\sqrt {10} }}}&{ - {3 \mathord{\left/

{\vphantom {3 {\sqrt {10} }}} \right.

\kern-\nulldelimiterspace} {\sqrt {10} }}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {10}}} \right.

\kern-\nulldelimiterspace} {10}}}&{{{ - 3} \mathord{\left/

{\vphantom {{ - 3} {10}}} \right.

\kern-\nulldelimiterspace} {10}}}\\{{{ - 3} \mathord{\left/

{\vphantom {{ - 3} {10}}} \right.

\kern-\nulldelimiterspace} {10}}}&{{9 \mathord{\left/

{\vphantom {9 {10}}} \right.

\kern-\nulldelimiterspace} {10}}}\end{aligned}} \right)\end{aligned}\)

Hence,\({U^T}U = \left( 1 \right)\)and\(U{U^T} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {10}}} \right.

\kern-\nulldelimiterspace} {10}}}&{{{ - 3} \mathord{\left/

{\vphantom {{ - 3} {10}}} \right.

\kern-\nulldelimiterspace} {10}}}\\{{{ - 3} \mathord{\left/

{\vphantom {{ - 3} {10}}} \right.

\kern-\nulldelimiterspace} {10}}}&{{9 \mathord{\left/

{\vphantom {9 {10}}} \right.

\kern-\nulldelimiterspace} {10}}}\end{aligned}} \right)\).

Write the Theorem

Let \({\bf{y}} \in {\mathbb{R}^n}\), then \({\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\), if \(U = \left( {\begin{aligned}{}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}& \cdots &{{{\bf{u}}_p}}\end{aligned}} \right)\).

Find \({\bf{Pro}}{{\bf{j}}_W}y\) and \(\left( {{U^T}U} \right)y\)

(b)

It is observed\(U = \left( {{{\bf{u}}_1}} \right)\)from part (a), then\({\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\).

From part (a), \(U{U^T} = \left( {\begin{aligned}{}{{1 \mathord{\left/{\vphantom {1 {10}}} \right.} {10}}}&{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}\\{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}&{{9 \mathord{\left/

{\vphantom {9 {10}}} \right} {10}}}\end{aligned}} \right)\), then find \(U{U^T}{\bf{y}}\), where \({\bf{y}} = \left( {\begin{aligned}{}7\\9\end{aligned}} \right)\).

\(\begin{aligned}{c}U{U^T}{\bf{y}} = \left( {\begin{aligned}{}{{1 \mathord{\left/

{\vphantom {1 {10}}} \right} {10}}}&{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}\\{{{ - 3} \mathord{\left/{\vphantom {{ - 3} {10}}} \right.} {10}}}&{{9 \mathord{\left/ {\vphantom {9 {10}}} \right.

\kern-\nulldelimiterspace} {10}}}\end{aligned}} \right)\left( {\begin{aligned}{}7\\9\end{aligned}} \right)\\ = \left( {\begin{aligned}{}{\frac{7}{{10}} - \frac{{27}}{{10}}}\\{ - \frac{{21}}{{10}} + \frac{{81}}{{10}}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{}{ - \frac{{20}}{{10}}}\\{\frac{{60}}{{10}}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{}{ - 2}\\6\end{aligned}} \right)\end{aligned}\)

So,\({\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = \left( {\begin{aligned}{}{ - 2}\\6\end{aligned}} \right)\), as\({\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\).