Page 431 - 35Linear Algebra

G.15 Kernel, Range, Nullity, Rank 431

n
n
corresponds to a linear transformation L: R → R , so condition 3 is equiva-
lent to condition 1.
Condition 6 implies 4 by the adjoint construct the inverse, but the con-
verse 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 eigen-
value 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. Show that ker L = {0 V } if and
only if L is one-to-one:
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 6= 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?

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