linear-guest

310 Least squares and Singular Values
As usual, our starting point is the computation of L acting on the input basis
vectors;
LO = ( Lu 1 , . . . , Lu n
)
=
(√
λ1 v 1 , . . . , √ λ n v n
)
⎛√ λ1 0 · · · 0
√ 0 λ2 · · · 0
= ( .
)
. . .. .
v 1 , . . . , v m 0 0 · · ·
0 0 · · · 0
⎜
⎝ . . .
0 0 · · · 0
√
λn
⎞
.
⎟
⎠
The result is very close to diagonalization; the numbers √ λ i along the leading
diagonal are called the singular values of L.
Example 155 Let the matrix of a linear transformation be
⎛ ⎞
1
2
1
2
⎜
M = ⎝−1 ⎟
1⎠ .
− 1 2
− 1 2
Clearly ker M = {0} while
M T M =
which has eigenvalues and eigenvectors
λ = 1 , u 1 :=
(
√2 1
)
√1
2
( 3
2
− 1 )
2
− 1 2
3
2
; λ = 2 , u 2 :=
(
√2 1
)
− √ 1
2
.
so our orthonormal input basis is
O =
((
√2 1
)
√1
2
,
(
√2 1
))
− √ 1
2
These are called the right singular vectors of M. The vectors
⎛ ⎞
⎛ ⎞
√1
0
⎜ 2
⎟
⎜
Mu 1 = ⎝ 0 ⎠ and Mu 2 = ⎝ − √ ⎟
2 ⎠
− √ 1
2 0
310
.