Linear Transformations on Rn
Definition of a Linear Transformation
In your travels throughout your mathematical career there has been one theme that persists in every course. That theme is functions. Recall that a function is a rule that assigns every element from a domain set to a unique element of a range set. If the domain and range are both the real numbers, then a function is the familiar real valued function. If the domain is a real number and the range is Rn then the function is a vector valued function or a parametrically defined curve. If the domain is Rn and the range is the real numbers, then the function is a function of several variables. In linear algebra we are interested in special functions where the domain is Rn and the range is Rm.
L: Rn ---> Rm
be a function such that the following two properties hold:
Then the function is called a linear transformation.
This definition calls for some examples.
L: R2 ---> R3
be defined by
L(x,y) = (y, x, x + y)
Show that L is a linear transformation.
First, we prove the first property. Suppose that
u = (u1,u2) and v = (v1,v2)
L(u + v) = L(u1 + v1, u2 + v2)
= (u2 + v2, u1 + v1, u1 + v1 + u2 + v2)
L(u) + L(v) = (u2, u1, u1 + u2) + (v2, v1, v1 + v2)
= (u2 + v2, u1 + v1, u1 + u2 +v1 + v2)
L(u + v) = L(u) + L(v)
Now for the second property. We have
L(cu) = L(c(u1, u2)) = L(cu1, cu2)
= (cu2, cu1, cu1 + cu2) = c(u2, u1, u1 + u2) = cL(u)
Since properties 1 and 2 hold, we can conclude that L is a linear transformation.
Show that the function
f: R3 ---> R2
f(x, y, z) = (xy, yz)
is not a linear transformation
To show a function is not a linear transformation, we just need to find an example that demonstrates the failure of one of the properties. We have
f(2(3, 4, 5)) = f(6, 8, 10) = (48, 80)
2(f(3, 4, 5)) = 2(12, 20) = (24, 40)
since these are not equal, we can conclude that f is not a linear transformation.
It is a simple consequence to the two properties that if L is a linear transformation then
L(c1v1 + c2v2 + ... + ckvk) = c1L(v1) + c2L(v2) + ... + ckL(vk)
or in sigma notation
L(Scivi) = S L(civi)
The Matrix of a Linear Transformation
Let A be an m x n matrix and let
L: Rn ---> Rm
be defined by
L(u) = AuT
Then L is a linear transformation.
The proof is just a matter of stating the corresponding properties of matrices. We have
L(u + v) = A(u + v)T = A(uT + vT)
= AuT + AvT = L(u) + L(v)
L(cu) = A(cu)T = A(cuT)
= cAuT = cL(u)
The converse is also true.
L: Rm ---> Rn
be a linear transformation. Then there is a unique matrix A such that
L(u) = AuT
Recall that the vector ei is the vector with ith component equal to 1 and all others zero. Any vector
v = (v1,v2, ... ,vm) = v1e1 + v2e2 + ... + vmem
We let the ith column of A be the vector
Aei = the ith column of A
L(ei) = Aei
L(v) = L(v1, v2, ... , vm) = L(v1e1 + v2e2 + ... + vmem)
= v1L(e1) + v2L(e2) + ... + vmL(em) = v1Ae1 + v2Ae2 + ... + vmAem
= A(v1e1 + v2e2 + ... + vmem) = AvT
To show that the matrix is unique, we just notice that if B is a matrix with
L(v) = BvT
Bei = L(ei) = Aei
so that the columns of A and B are the same.
For the linear transformation from the first example,
L(x, y) = (y, x, x + y)
L(1, 0) = (0, 1, 1) L(0, 1) = (1, 0, 1)
so that the matrix A that represents the linear transformation is
Properties of Linear Transformations
There are a few notable properties of linear transformation that are especially useful. They are the following.
Notice that in the first property, the 0's on the left and right hand side are different. The left hand 0 is the zero vector in Rm and the right hand 0 is the zero vector in Rn. The proofs of these can be done in two ways. One way is to use the definition of a linear transformation. A quicker way is to note that a linear transformation can be represented as a matrix. These two properties are just properties of matrices.