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Linear Algebra Exercises-n-Answers.pdf

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<strong>Answers</strong> to <strong>Exercises</strong> 131<br />

Three.VI.1.21<br />

the projections are these.<br />

The sequence need not settle down. With<br />

( ( )<br />

1<br />

⃗a = ⃗ 1 b =<br />

0)<br />

1<br />

⃗v 1 =<br />

This sequence doesn’t repeat.<br />

( )<br />

1/2<br />

, ⃗v<br />

1/2 2 =<br />

( )<br />

1/2<br />

, ⃗v<br />

0 3 =<br />

( )<br />

1/4<br />

, . . .<br />

1/4<br />

Subsection Three.VI.2: Gram-Schmidt Orthogonalization<br />

Three.VI.2.9 (a)<br />

(<br />

1<br />

⃗κ 1 =<br />

1)<br />

( ) (<br />

2 1<br />

( ( ( (<br />

2 2 2 1 1)<br />

1<br />

⃗κ 2 = − proj<br />

1)<br />

[⃗κ1]( ) = − ) ( · =<br />

1)<br />

1)<br />

1 1)<br />

1)<br />

(<br />

1<br />

1<br />

(<br />

2<br />

−<br />

1)<br />

3 ( ( )<br />

1 1/2<br />

2 1)<br />

· =<br />

−1/2<br />

(b)<br />

(<br />

0<br />

⃗κ 1 =<br />

1)<br />

⃗κ 2 =<br />

( )<br />

−1<br />

− proj<br />

3<br />

[⃗κ1 ](<br />

(<br />

−1<br />

3<br />

)<br />

) =<br />

( )<br />

−1<br />

−<br />

3<br />

( ) (<br />

−1 0<br />

(<br />

3 1)<br />

0<br />

( ) ( · =<br />

0 0 1)<br />

1 1)<br />

( )<br />

−1<br />

− 3 (<br />

0<br />

3 1 1)<br />

· =<br />

( )<br />

−1<br />

0<br />

(c)<br />

(<br />

0<br />

⃗κ 1 =<br />

1)<br />

⃗κ 2 =<br />

( )<br />

−1<br />

− proj<br />

0<br />

[⃗κ1 ](<br />

(<br />

−1<br />

0<br />

)<br />

) =<br />

( )<br />

−1<br />

−<br />

0<br />

( ) ( −1 0<br />

(<br />

0 1)<br />

0<br />

( ) ( · =<br />

0 0 1)<br />

1 1)<br />

( )<br />

−1<br />

− 0 (<br />

0<br />

0 1 1)<br />

· =<br />

( )<br />

−1<br />

0<br />

The corresponding orthonormal bases for the three parts of this question are these.<br />

( √ ) ( √ ) ( ( ) ( ( )<br />

1/ 2<br />

〈<br />

1/ √ 2/2<br />

,<br />

2 − √ 0 −1 0 −1<br />

〉 〈 , 〉 〈 , 〉<br />

2/2 1)<br />

0 1)<br />

0

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