linear-guest

G.15 Kernel, Range, Nullity, Rank 431
corresponds to a linear transformation L: R n → R n , so condition 3 is equivalent
to condition 1.
Condition 6 implies 4 by the adjoint construct the inverse, but the converse
is not so obvious. For the converse (4 implying 6), we refer back the
proofs in Chapter 18 and 19. Note that if det M = 0, there exists an eigenvalue
of M equal to 0, which implies M is not invertible. Thus condition 8
is equivalent to conditions 4, 5, 9, and 10.
The map M is injective if it does not have a null space by definition,
however eigenvectors with eigenvalue 0 form a basis for the null space. Hence
conditions 8 and 14 are equivalent, and 14, 15, and 16 are equivalent by the
Dimension Formula (also known as the Rank-Nullity Theorem).
Now conditions 11, 12, and 13 are all equivalent by the definition of a
basis. Finally if a matrix M is not row-equivalent to the identity matrix,
then det M = 0, so conditions 2 and 8 are equivalent.
Hint for Review Problem 2
Lets work through this problem.
Let L: V → W be a linear transformation.
only if L is one-to-one:
Show that ker L = {0 V } if and
1. First, suppose that ker L = {0 V }. Show that L is one-to-one.
Remember what one-one means, it means whenever L(x) = L(y) we can be
certain that x = y. While this might seem like a weird thing to require
this statement really means that each vector in the range gets mapped to
a unique vector in the range.
We know we have the one-one property, but we also don’t want to forget
some of the more basic properties of linear transformations namely that
they are linear, which means L(ax + by) = aL(x) + bL(y) for scalars a and
b.
What if we rephrase the one-one property to say whenever L(x) − L(y) = 0
implies that x − y = 0? Can we connect that to the statement that ker L =
{0 V }? Remember that if L(v) = 0 then v ∈ ker L = {0 V }.
2. Now, suppose that L is one-to-one. Show that ker L = {0 V }. That is, show
that 0 V is in ker L, and then show that there are no other vectors in
ker L.
What would happen if we had a nonzero kernel? If we had some vector v
with L(v) = 0 and v ≠ 0, we could try to show that this would contradict
the given that L is one-one. If we found x and y with L(x) = L(y), then
we know x = y. But if L(v) = 0 then L(x) + L(v) = L(y). Does this cause a
problem?
431