Finding dimension of kernel of linear transformation
Answers
The Kernel and the Range of a Linear Transformation
One to One Linear Transformations
Recall that a function is 1-1 if
f(x) = f(y)
implies that
x = y
Since a linear transformation is defined as a function, the definition of 1-1 carries over to linear transformations. That is
Definition
A linear transformation L is 1-1 if for all vectors u and v,
L(u) = L(v)
implies that
u = v
Example
Let L be the linear transformation from R2 to P2 defined by
L((x,y)) = xt2 + yt
We can verify that L is indeed a linear transformation. We now check that L is 1-1. Let
L(x1,y1) = L(x2,y2)
then
x1t2 + y1t = x2t2 + y2t
If two polynomials are equal to each other, then their coefficients are all equal. In particular,
x1 = x2 and y1 = y2
We can conclude that L is a 1-1 linear transformation.
Example
Let L be the linear transformation from P1 to R1 defined by
L(f(t)) = f(0)
Then L is not a 1-1 linear transformation since
L(0) = L(t)
and
0 1
The Kernel
Related to 1-1 linear transformations is the idea of the kernel of a linear transformation.
Definition
The kernel of a linear transformation L is the set of all vectors v such that
L(v) = 0
Example
Let L be the linear transformation from M2x2 to P1 defined by
Then to find the kernel of L, we set
(a + d) + (b + c)t = 0
d = -a c = -b
so that the kernel of L is the set of all matrices of the form
Notice that this set is a subspace of M2x2.
Theorem
The kernel of a linear transformation from a vector space V to a vector space W is a subspace of V.
Proof
Suppose that u and v are vectors in the kernel of L. Then
L(u) = L(v) = 0
We have
L(u + v) = L(u) + (v) = 0 + 0 = 0
and
L(cu) = cL(u) = c 0 = 0
Hence u + v and cu are in the kernel of L. We can conclude that the kernel of L is a subspace of V.
In light of the above theorem, it makes sense to ask for a basis for the kernel of a linear transformation. In the previous example, a basis for the kernel is given by
Next we show the relationship between 1-1 linear transformations and the kernel.
Theorem
A linear transformation L is 1-1 if and only if Ker(L) = 0.
Proof
Let L be 1-1 and let v be in Ker(L). We need to show that v is the zero vector. We have both
L(v) = 0 and L(0) = 0
Since L is 1-1,
v = 0
Now let Ker(L) = 0. Then
L(u) = L(v)
implies that
0 = L(v) - L(u) = L(v - u)
Hence v - u is in Ker(L), so that
v - u = 0 or u = v
and L is 1-1.
Range
We have seen that a linear transformation from V to W defines a special subspace of V called the kernel of L. Now we turn to a special subspace of W.
Definition
Let L be a linear transformation from a vector space V to a vector space W. Then the range of L is the set of all vectors w in W such that there is a v in V with
L(v) = w
Theorem
The range of a linear transformation L from V to W is a subspace of W.
Proof
Let w1 and w2 vectors in the range of W. Then there are vectors v1 and v2 with
L(v1) = w1 and L(v2) = w2
We must show closure under addition and scalar multiplication. We have
L(v1 + v2) = L(v1) + L(v2) = w1 + w2
and
L(cv1) = cL(v1) = cw1
hence w1 + w2 and cw1 are in the range of L. Hence the range of L is a subspace of W.
We say that a linear transformation is onto W if the range of L is equal to W.
Example
Let L be the linear transformation from R2 to R3 defined by
L(v) = Av
with
A. Find a basis for Ker(L).
B. Determine of L is 1-1.
C. Find a basis for the range of L.
D. Determine if L is onto.