Page 292 - 35Linear Algebra
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292 Kernel, Range, Nullity, Rank
16.2.2 Kernel
Let L: V → W be a linear transformation. Suppose L is not injective. Then
we can find v 1 6= v 2 such that Lv 1 = Lv 2 . So v 1 − v 2 6= 0, but
L(v 1 − v 2 ) = 0.
Definition If L: V → W is a linear function then the set
ker L = {v ∈ V | Lv = 0 W } ⊂ V
is called the kernel of L.
Notice that if L has matrix M in some basis, then finding the kernel of L
is equivalent to solving the homogeneous system
MX = 0.
Example 147 Let L(x, y) = (x + y, x + 2y, y). Is L one-to-one?
To find out, we can solve the linear system:
1 1 0 1 0 0
1 2 0 ∼ 0 1 0 .
0 1 0 0 0 0
Then all solutions of MX = 0 are of the form x = y = 0. In other words, ker L = {0},
and so L is injective.
Reading homework: problem 1
Notice that in the above example we found
1 1 1 0
ker 1 2 = ker 0 1 .
0 1 0 0
In general, an efficient way to get the kernel of a matrix is to write a string
of equalities between kernels of matrices which differ by row operations and,
once RREF is reached, note that the linear relationships between the columns
for a basis for the nullspace.
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