Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
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<strong>Answers</strong> to <strong>Exercises</strong> 135<br />
the Gram-Schmidt⎛process.<br />
⎞<br />
⃗κ 1 =<br />
⎝ 1 5 ⎠<br />
−1<br />
⎛<br />
⃗κ 2 = ⎝ 2 ⎞ ⎛<br />
2⎠ − proj [⃗κ1 ]( ⎝ 2 ⎞ ⎛<br />
2⎠) = ⎝ 2 ⎞ ⎝ 2 2⎠<br />
⎝ 1 5 ⎠ ⎛<br />
0 −1<br />
2⎠ − ⎛<br />
0<br />
0 0<br />
⎝ 1 ⎞ ⎛<br />
5 ⎠ ⎝ 1 ⎞ · ⎝ 1 ⎞<br />
5 ⎠<br />
−1<br />
5 ⎠<br />
−1 −1<br />
⎛<br />
= ⎝ 2 ⎞ ⎛<br />
2⎠ − 12<br />
27 · ⎝ 1 ⎞ ⎛<br />
5 ⎠ = ⎝ 14/9<br />
⎞<br />
−2/9⎠<br />
0 −1 4/9<br />
⎛ ⎞ ⎛ ⎞<br />
⎛ ⎞<br />
⃗κ 3 = ⎝ 1 0⎠ − proj [⃗κ1 ]( ⎝ 1 0⎠) − proj [⃗κ2 ]( ⎝ 1 0⎠)<br />
0<br />
0<br />
0<br />
⎛<br />
⎛<br />
= ⎝ 1 ⎞ ⎝ 1 ⎞ ⎛<br />
0⎠<br />
⎝ 1 ⎞<br />
⎛<br />
5 ⎠ ⎛<br />
0 −1<br />
0⎠ − ⎛<br />
0<br />
⎝ 1 ⎞ ⎛<br />
5 ⎠ ⎝ 1 ⎞ · ⎝ 1 ⎞ ⎝ 1 ⎞ ⎛<br />
0⎠<br />
⎝ 14/9<br />
⎞<br />
−2/9⎠<br />
⎛<br />
0 4/9<br />
5 ⎠ − ⎛<br />
−1<br />
5 ⎠<br />
⎝ 14/9<br />
⎞ ⎛<br />
−2/9⎠<br />
⎝ 14/9<br />
⎞ · ⎝ 14/9<br />
⎞<br />
−2/9⎠<br />
4/9<br />
−2/9⎠<br />
−1 −1<br />
4/9 4/9<br />
⎛<br />
= ⎝ 1 ⎞ ⎛<br />
0⎠ − 1<br />
27 · ⎝ 1 ⎞ ⎛<br />
5 ⎠ − 7 12 · ⎝ 14/9<br />
⎞ ⎛<br />
−2/9⎠ = ⎝ 1/18<br />
⎞<br />
−1/18⎠<br />
0 −1<br />
4/9 −4/18<br />
The result ⃗κ 3 is orthogonal to both ⃗κ 1 and ⃗κ 2 . It is therefore orthogonal to every vector in the span<br />
of the set {⃗κ 1 , ⃗κ 2 }, including the two vectors given in the question.<br />
Three.VI.2.17<br />
(a) The<br />
(<br />
representation<br />
(<br />
can be done<br />
(<br />
by eye.<br />
2 1 1<br />
= 3 · + (−1) · Rep<br />
3)<br />
1)<br />
0)<br />
B (⃗v ) =<br />
The two projections ( are ) also ( easy.<br />
2 1<br />
( )<br />
( 2<br />
proj [ β1 ⃗ ] ( 3 1)<br />
1<br />
) = ) ( ·<br />
3<br />
1 1)<br />
1)<br />
(<br />
1<br />
1<br />
= 5 ( 1<br />
2 1)<br />
·<br />
⎛<br />
⎞<br />
⎛<br />
⎞<br />
( ) 2<br />
proj [ β2 ⃗ ] ( ) = (<br />
3 1<br />
0<br />
( )<br />
3<br />
−1<br />
B<br />
( ) (<br />
2 1<br />
3 0)<br />
) ( · 1<br />
0)<br />
( 1<br />
0)<br />
= 2 ( 1<br />
1 0)<br />
·<br />
(b) As above, the representation<br />
(<br />
can be done by eye<br />
2<br />
= (5/2) ·<br />
3)<br />
and the two projections ( ) ( are easy.<br />
2 1<br />
( )<br />
(<br />
2<br />
proj [ β1 ⃗ ] ( 3 1)<br />
1<br />
) = ) ( ·<br />
3<br />
1 1)<br />
1)<br />
= 5 (<br />
1<br />
2 1)<br />
·<br />
(<br />
1<br />
1<br />
Note the recurrence of the 5/2 and the −1/2.<br />
(c) Represent ⃗v with respect to the basis<br />
(<br />
1<br />
1)<br />
+ (−1/2) ·<br />
Rep K (⃗v ) =<br />
( )<br />
1<br />
−1<br />
( ) (<br />
2 1<br />
( )<br />
( )<br />
2<br />
proj [ β2 ⃗ ] ( 3 −1)<br />
1<br />
) = ( ) ( · =<br />
3 1 1 −1<br />
−1 −1) −1 ·( )<br />
1<br />
2 0<br />
⎛<br />
⎜<br />
⎝<br />
⎞<br />
r 1<br />
⎟<br />
. ⎠<br />
r k<br />
so that ⃗v = r 1 ⃗κ 1 + · · · + r k ⃗κ k . To determine r i , take the dot product of both sides with ⃗κ i .<br />
⃗v ⃗κ i = (r 1 ⃗κ 1 + · · · + r k ⃗κ k ) ⃗κ i = r 1 · 0 + · · · + r i · (⃗κ i ⃗κ i ) + · · · + r k · 0<br />
Solving for r i yields the desired coefficient.<br />
(d) This is a restatement of the prior item.<br />
Three.VI.2.18 First, ‖⃗v ‖ 2 = 4 2 + 3 2 + 2 2 + 1 2 = 50.
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