Page 306 - 35Linear Algebra

306 Least squares and Singular Values

Then he should find a and b to best fit the data.

11 = a · 1 + b
19 = a · 2 + b
31 = a · 3 + b.

As a system of linear equations, this becomes:

   
1 1 11
a ?
2 1 = 19 .
   
b
3 1 31
T
T
There is likely no actual straight line solution, so instead solve M MX = M V .
   
1 1 11

1 2 3 a 1 2 3
 2 1  =  19  .
1 1 1 b 1 1 1
3 1 31
This simplifies to

14 6 142 1 0 10
∼ 1 .
6 3 61 0 1
3
Thus, the least-squares fit is the line
1
v = 10 t + .
3
Notice that this equation implies that Captain Conundrum accelerates towards Italian
2
soil at 10 m/s (which is an excellent approximation to reality) and that he started at
a downward velocity of 1 m/s (perhaps somebody gave him a shove...)!
3

17.1 Projection Matrices

T
T
We have seen that even if MX = V has no solutions M MX = M V does
have solutions. One way to think about this is, since the codomain of M is
the direct sum
codom M = ranM ⊕ ker M T
T
there is a unique way to write V = V r +V k with V k ∈ ker M and V r ∈ ran M,
and it is clear that Mx = V only has a solution of V ∈ ran M ⇔ V k = 0.
If not, then the closest thing to a solution of MX = V is a solution to

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