Linear Algebra, 2020a

Section VI. Projection 281
For the second member of the new basis, we subtract from ⃗β 2 the part in the
direction of ⃗κ 1 . This leaves the part of ⃗β 2 that is orthogonal to ⃗κ 1 .
⃗κ 2 =
( ( ( ( 1 1 1 2
− proj
3)
[⃗κ1 ]( )= − =
3)
3)
1)
( ) ⃗κ −1
2
2
By the corollary 〈⃗κ 1 ,⃗κ 2 〉 is a basis for R 2 .
2.5 Definition An orthogonal basis for a vector space is a basis of mutually
orthogonal vectors.
2.6 Example To produce from this basis for R 3
⎛
⎜
1
⎞ ⎛
⎟ ⎜
0
⎞ ⎛
⎟ ⎜
1
⎞
⎟
B = 〈 ⎝1⎠ , ⎝2⎠ , ⎝0⎠〉
1 0 3
an orthogonal basis, start by taking the first vector unchanged.
⎛
⎜
1
⎞
⎟
⃗κ 1 = ⎝1⎠
1
Get ⃗κ 2 by subtracting from ⃗β 2 its part in the direction of ⃗κ 1 .
⎛ ⎞ ⎛ ⎞ ⎛
0
0
⎜ ⎟ ⎜ ⎟ ⎜
0
⎞ ⎛
⎟ ⎜
2/3
⎞ ⎛ ⎞
−2/3
⎟ ⎜ ⎟
⃗κ 2 = ⎝2⎠ − proj [⃗κ1 ]( ⎝2⎠) = ⎝2⎠ − ⎝2/3⎠ = ⎝ 4/3 ⎠
0
0 0 2/3 −2/3
Find ⃗κ 3 by subtracting from ⃗β 3 the part in the direction of ⃗κ 1 and also the part
in the direction of ⃗κ 2 .
⎛ ⎞ ⎛ ⎞
⎛
1
1
⎜ ⎟ ⎜ ⎟
⎜
1
⎞ ⎛ ⎞
−1
⎟ ⎜ ⎟
⃗κ 3 = ⎝0⎠ − proj [⃗κ1 ]( ⎝0⎠)−proj [⃗κ2 ]( ⎝0⎠) = ⎝ 0 ⎠
3
3
3 1
As above, the corollary gives that the result is a basis for R 3 .
⎛
⎜
1
⎞ ⎛
⎟ ⎜
−2/3
⎞ ⎛
⎟ ⎜
−1
⎞
⎟
〈 ⎝1⎠ , ⎝ 4/3 ⎠ , ⎝ 0 ⎠〉
1 −2/3 1