Page 432 - 35Linear Algebra

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G.16 Least Squares and Singular Values

Least Squares: Hint for Review Problem 1
Lets work through this problem. Let L : U → V be a linear transformation.
Suppose v ∈ L(U) and you have found a vector u ps that obeys L(u ps ) = v.
Explain why you need to compute ker L to describe the solution space of the
linear system L(u) = v.
Remember the property of linearity that comes along with any linear trans-
formation: L(ax + by) = aL(x) + bL(y) for scalars a and b. This allows us to
break apart and recombine terms inside the transformation.
Now suppose we have a solution x where L(x) = v. If we have an vector
y ∈ ker L then we know L(y) = 0. If we add the equations together L(x) + L(y) =
L(x + y) = v + 0 we get another solution for free. Now we have two solutions,
is that all?

Hint for Review Problem 2

For the first part, what is the transpose of a 1 × 1 matrix? For the other two
T
parts, note that v v = v v. Can you express this in terms of kvk? Also you
need the trivial kernel only for the last part and just think about the null
space of M. It might help to substitute w = Mx.

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