Note to Student: In this section we will often use , and to denote subspaces of , or any other finite-dimensional vector space, such as those we study in Vector Spaces.
Image and Kernel of a Linear Transformation
The Image of a Linear Transformation
Let and be vector spaces, and let be a linear transformation. The image of ,
denoted by , is the set
In other words, the image of consists of individual images of all vectors of
.
Consider the linear transformation with standard matrix
item:impart1 Let then
Thus, every element of the image can be written as a linear combination of the columns of . We conclude that
Every column of is a scalar multiple of . Thus,
The image of is a line in determined by the vector .
item:impart2 The action of can be illustrated with a sketch.
In Example ex:image1 we observed that the image of the linear transformation was equal to the column space of its standard matrix. In general, it is easy to see that if is a linear transformation with standard matrix then the following relationship holds: In addition, by Theorem th:dimroweqdimcoleqrank, we know that
Let be a linear transformation with standard matrix
Find and .
As in Example ex:image1, the image of is given by
This time it is harder to detect the vectors that can be eliminated from the spanning
set without affecting the span. We have to rely on the reduced row-echelon form of
.
We can see that , so .
To identify vectors that span , we turn to Procedure proc:colspace. We identify the first three columns as pivot columns. These columns are linearly independent and span . Therefore,
By Theorem th:span_is_subspace and Definition def:colspace, we know that for an matrix , is a subspace of . However, when vector spaces other than are involved, it is not yet clear that is a subspace of the codomain. The following theorem resolves this issue.
- Proof
- To show that is a subspace, we need to show that is closed under
addition and scalar multiplication.
Suppose and are in . Then there are vectors and in such that and . Then This shows that is in .
For any scalar , we have: This shows that is in .
We can now define the rank of a linear transformation.
This definition gives us the following relationship between the rank of a linear transformation and the rank of the standard matrix associated with it.
The Kernel of a Linear Transformation
Let and be vector spaces, and let be a linear transformation. The kernel of ,
denoted by , is the set
In other words, the kernel of consists of all vectors of that map to in
.
It is important to pay attention to the locations of the kernel and the image. We
already proved that is a subspace of the codomain. In contrast, is located in the
domain. (We will prove shortly that it is a subspace of the domain.)
Let be a linear transformation with standard matrix
item:kernelT To find the kernel of , we need to find all vectors of that map to in . This amounts
to solving the equation .
Gauss-Jordan elimination yields:
Thus, the kernel of consists of all elements of the form:
We conclude that
item:dimkernelT Since is the span of two vectors of , we know that is a subspace of . (See Theorem th:span_is_subspace.) Observe that the two vectors in the spanning set are linearly independent. (How can we see this without performing computations?) Therefore .
Recall that the null space of a matrix is defined to be set of all solutions to the homogeneous equation . This means that if is a linear transformation with standard matrix then
We know that of an matrix is a subspace of . (See Theorem th:nullsubspacern.) We conclude this section by showing that even when vector spaces other than are involved, the kernel of a linear transformation is a subspace of the domain of the transformation.
- Proof
- To show that is a subspace, we need to show that is closed under
addition and scalar multiplication.
Suppose that and are in . Then, This shows that is in .
For any scalar we have: This shows that is in .
This definition gives us the following relationship between nullity of a linear transformation and the nullity of the standard matrix associated with it.
Rank-Nullity Theorem for Linear Transformations
In Examples ex:image2 and ex:kernel, we found the image and the kernel of the linear transformation with standard matrix
We also found that and
Because of Rank-Nullity Theorem for matrices (Theorem th:matrixranknullity), it is not surprising that
The following theorem is a generalization of this result.
- Proof
- By Theorem th:imagesubspace, is a subspace of . There exists a basis for of the form .
By Theorem th:kersubspace, is a subspace of . Let be a basis for .
We will show that is a basis for .
For any vector in , we have: for some scalars . Thus, By linearity, Therefore is in .
Hence there are scalars such that Thus,
We conclude that
Now we need to show that is linearly independent.
Suppose
Applying to both sides, we getBut for , thus Since is linearly independent, it follows that each .
But then Equation (eq:kerplusimproof) implies that . Because is linearly independent, it follows that each .
We conclude that is a basis for . Thus,
Practice Problems
Describe the image and find the rank for each linear transformation with standard
matrix given below.
Suppose linear transformations and are such that . Does this mean that and are
the same transformation? Justify your claim.