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Since
proj
L
(
~v
2
)=
1
k
~v
1
k
2
(
~v
2
·
~v
1
)
~v
1
=
5
3
~v
1
=
5
3
~v
1
+0
~v
2
,
we have
[proj
L
(
~v
2
)]
{
~v
1
,~v
2
}
=
µ
5
/
3
0
¶
(b) Since
proj
L
(
1
~v
1
,
proj
L
(
~v
2
5
3
~v
1
,
we conclude that the matrix of proj
L
in the basis
~v
1
,~v
2
is
[proj
L
]
{
~v
1
,~v
2
}
=
µ
15
/
3
00
¶
6. Let us Frst “orhonormalize” the vectors
~v
1
=
⎛
⎝
2
1
2
⎞
⎠
,~
v
2
=
⎛
⎝
0
3
3
⎞
⎠
,
i.e. Fnd an orthonormal basis of the plane. The GramSchmidt process gives
~u
1
=
⎛
⎝
2
/
3
1
/
3
2
/
3
⎞
⎠
u
2
=
⎛
⎝
−
2
/
3
2
/
3
1
/
3
⎞
⎠
.
±inally we use the formula:
proj
span(
~v
1
,~v
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This note was uploaded on 04/12/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.
 Fall '08
 lee

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