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Chapter 1: Q36E (page 1)

Let \(H = Span\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}}} \right\}\) and \(K = Span\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}}} \right\}\), where

\({u_1} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{0}}\\{ - {\bf{1}}}\end{array}} \right)\),\({u_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{2}}\\{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right)\),\({u_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{4}}\\{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right)\),

\({v_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right)\),\({v_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{2}}\\{ - {\bf{6}}}\end{array}} \right)\),\({v_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{4}}\\{\bf{6}}\\{ - {\bf{2}}}\end{array}} \right)\)

Find bases for H, K, and \(H + K\).

Short Answer

Expert verified

The set \(\left\{ {u{ & _1},{u_2}} \right\}\) is a basis for H, \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is a basis for K, and \(\left\{ {{u_1},{u_2},{v_2},{v_3}} \right\}\) is a basis for \(H + K\).

Step by step solution

01

Use the spanning set theorem

By the spanning set theorem, the subset of \(\left\{ {{u_1},{u_2},{u_3}} \right\}\) forms a basis for H. Similarly, the subset of \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) forms a basis for K.

02

Compute the basis for H

Use the row-reduced echelon form to identify the pivot columns.

\(\left( {\begin{array}{*{20}{c}}{{u_1}}&{{u_2}}&{{u_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&3\\2&2&4\\0&{ - 1}&1\\{ - 1}&1&{ - 4}\end{array}} \right)\)

Add \( - 2\) times row 1 to row 2, and add row 1 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3\\0&2&{ - 2}\\0&{ - 1}&1\\0&1&{ - 1}\end{array}} \right)\)

Divide row 2 by 2, and add row 3 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3\\0&1&{ - 1}\\0&{ - 1}&1\\0&0&0\end{array}} \right)\)

Now, add rows 2 and 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3\\0&1&{ - 1}\\0&0&0\\0&0&0\end{array}} \right)\)

Thus \(\left\{ {u{ & _1},{u_2}} \right\}\) is a basis for H.

03

Compute the basis for K

Use the row-reduced echelon form to identify the pivot columns.

\(\left( {\begin{array}{*{20}{c}}{{v_1}}&{{v_2}}&{{v_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}&2&{ - 1}\\{ - 2}&3&4\\{ - 1}&2&6\\3&{ - 6}&{ - 2}\end{array}} \right)\)

Divide row 1 by \( - 2\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 1}&{0.5}\\{ - 2}&3&4\\{ - 1}&2&6\\3&{ - 6}&{ - 2}\end{array}} \right)\)

Add 2 times row 1 to row . Then add 1 time row 1 to row 3.

Again add \( - 3\) times row 1 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 1}&{0.5}\\0&1&5\\0&1&{6.5}\\0&{ - 3}&{ - 3.5}\end{array}} \right)\)

Add \( - 1\) time row 2 to row 3, and add 3 times row 2 to row 3 .

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - 1}&{0.5}\\0&1&5\\0&0&{1.5}\\0&0&{ - 11.5}\end{array}} \right)\)

Finally, add row 2 to row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{5.5}\\0&1&5\\0&0&{1.5}\\0&0&{ - 11.5}\end{array}} \right)\)

Divide row 3 by \(1.5\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{5.5}\\0&1&5\\0&0&1\\0&0&{ - 11.5}\end{array}} \right)\)

Add \( - 5.5\) times row 3 to row 1. Then add \( - 5\) times row 3 to row 2. Again add \(11.5\) times row 3 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}} \right)\)

Thus, \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is a basis for K.

04

Compute the basis for \(H + K\)

Note that \(H + K = {\rm{Span}}\left\{ {{u_1},{u_2},{u_3},{v_1},{v_2},{v_3}} \right\}\).

Use the row-reduced echelon form to identify the pivot columns.

\(\left( {\begin{array}{*{20}{c}}{{u_1}}&{{u_2}}&{{u_3}}&{{v_1}}&{{v_2}}&{{v_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&2&{ - 1}\\2&2&4&{ - 2}&3&4\\0&{ - 1}&1&{ - 1}&2&6\\{ - 1}&1&{ - 4}&3&{ - 6}&{ - 2}\end{array}} \right)\)

Add \( - 2\) times row 1 to row 2, and add 1 time row 1 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&2&{ - 1}\\0&2&{ - 2}&2&{ - 1}&6\\0&{ - 1}&1&{ - 1}&2&6\\0&1&{ - 1}&1&{ - 4}&{ - 3}\end{array}} \right)\)

Divide row 2 by 2.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&2&{ - 1}\\0&1&{ - 1}&1&{ - 0.5}&3\\0&{ - 1}&1&{ - 1}&2&6\\0&1&{ - 1}&1&{ - 4}&{ - 3}\end{array}} \right)\)

Add 1 time row 2 to row 3, and \( - 1\) time row 2 to row 4.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&2&{ - 1}\\0&1&{ - 1}&1&{ - 0.5}&3\\0&0&0&0&{1.5}&9\\0&0&0&0&{ - 3.5}&{ - 6}\end{array}} \right)\)

Divide row 3 by \(1.5\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&2&{ - 1}\\0&1&{ - 1}&1&{ - 0.5}&3\\0&0&0&0&1&6\\0&0&0&0&{ - 3.5}&{ - 6}\end{array}} \right)\)

Add \(3.5\) times row 3 to row 4. Then add \(0.5\) time row 3 to row 2. Again add \( - 2\) times row 3 to row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&0&{ - 13}\\0&1&{ - 1}&1&0&6\\0&0&0&0&1&6\\0&0&0&0&0&{15}\end{array}} \right)\)

Divide row 4 by 15.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&0&{ - 13}\\0&1&{ - 1}&1&0&6\\0&0&0&0&1&6\\0&0&0&0&0&1\end{array}} \right)\)

Add \( - 6\) times row 4 to row 3. Then add \( - 6\) times row 4 to row 2. Again add 13 times row 4 to row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&3&{ - 2}&0&0\\0&1&{ - 1}&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{array}} \right)\)

Thus, \(\left\{ {{u_1},{u_2},{v_2},{v_3}} \right\}\) is a basis for \(H + K\).

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