IV Image and Kernel of Linear Transformations
Motivation:
In the last class, we looked at the linearity of this function:
F : R3  R2
F(x,y,z)=(x+y+z,2x-3y+4z)
How does a 3 dimensional space get ‘mapped into’ a 2 dimensional space? At least one
dimension ‘collapses’, or disappears.
Image
Let F : Rn  Rm be a linear mapping.
The image of the linear transformation F (Im F) is the set of image points in Rm. That is,

Im F = u  R m | there exists v  R n such that F v   u

Examples

non-linear example:
f(x) = sin(x)
the image is the interval (-1,1)
Consider F: R2  R2 given by F(x,y)=(-y,x). (rotation by 90 degrees)
Im F = entire plane, every point in R2.
Choose an arbitrary point (a,b), then the point (b,-a) will map there.

Consider F: R2  R2 given by F(x,y)=(x,x).
This is the projection onto the line x=y.
Im F = single line in R2 (the line x=y).
Choose an arbitrary point (a,a) on the line, then any point (a,*) will map there.

Consider F: R2  R2 given by F(x,y)=(0,0).
This is a map to the zero vector.
Im F = single point in R2 (the point (0,0)).

Consider F: R2  R3 given by F(x,y)=(2x,2y,x-2y).
Note that a two-dimensional space has been mapped into a 3 dimensional space.

Theorem
The image of a linear transformation T: Rn  Rm is the span of the images of a set of
basis vectors in Rn. In particular, if T(x)=Ax the image is the span of the column vectors
of A. (Note the dimension of the span is equal to the rank of A).
Proof: We need to prove that the span of the images of a set of basis vectors in Rn span
the entire image of T. In other words that any uIm F is in their span.
Suppose e1,e1,…,en form a standard basis for Rn and T(e1), T(e2) … ,T(en) denotes their
images. Note that A is constructed as A= T (e1 ) T (e2 ) ... T (en ) , so the columns of A
are the images of each basis vector. We choose a random vector uIm T. By definition of
being in the image, there exists a vector u Rn such that T(v)=u. Since e1,e1,…,en form a
standard basis, u=k1e1+k2e2+…+knen for some scalars ki.
By linearity, T(v)= T(k1e1+k2e2+…+knen)= k1T(e1)+k2 T(e2)+…+kn T(en) and thus
u=T(v) is in the span of T(e1), T(e2) … ,T(en).
Interpretation: We need only look at the images of a set of basis vectors to determine the
span of the image of a transformation. Each basis vector in Rn may generate a linearly
independent image vector, so that the dimension of the image is always  dimension of
the preimage.
Example
F: R2  R3 given by F(x,y)=(2x,2y,x-2y).
F(1,0)=(2,0,1).
F(0,1)=(0,2,-2).
These two vectors form a span for the image. The image is 2 dimensional, as expected.
image is spanned by { (2,0,1),(0,2,-2)}
after row reduction, image is spanned by { (1,0,.5),(0,1,-1)} and the image is 2D.
Example (#10)
F : R3  R2
F(x,y,z)=(x+y+z,2x-3y+4z)
Let’s look at the map of the basis vectors:
F(1,0,0)=(1,2)
F(0,1,0)=(1,-3)
F(0,0,1)=(1,4)
The image = span { (1,2),(1,-3),(1,4)}
After row reduction,
the image = span ( (1,0),(0.1))
A basis of the image = (1,0),(0,1) and it has
dimension 2.
Analysis:
The first two basis vectors map to linearly independent image vectors! By considering all
linear combinations, we can reach every point in R2. But the third basis vector maps to a
linearly dependent image vector, and thus overlays points on the same set.
(For example F(2,2,0)=(4,-2) but also F(9,0,-5)=(4,-2).)
Kernel
Let F : Rn  Rm be a linear mapping. The kernel of F, written Ker F, is the set of
elements in Rn that map into the zero vector in Rm.
In particular, if F(x)=Ax the kernel of F is the set of solutions to Ax=0.
Example
F : R3  R2
F(x,y,z)=(x+y+z,2x-3y+4z)
The kernel will be the set of solutions to F(x,y,z)=0.
Solve:
(#10) F(x,y,z)=(x+y+z,2x-3y+4z) = (0,0,0)
1 1 1 0 1 0 7 5 0


~
2  3 4 0 0 1  2 5 0
 x  0  7 
 y   0  s  2
     
  7 
The kernel is the set S= s    and has dimension 1.
  2 
Note: The dimension of the image was 2, and the dimension of the kernel was 1, which
adds up to 3.
Theorem
The image and the kernel of a linear transformation are subspaces and their dimensions
sum to n.
This is just another statement of the rank nullity theorem.
In detail:
A transformation can be described as Ax=T.
Rank = num of linearly independent column vectors of A = dimension of the image
(since the span of the columns of A is the image of T)
nullity = num of columns that don’t have pivots = dimension of the kernel
(since each column without a pivot generates a free variable
in the solution to Ax=0)