We define the image and kernel of a linear transformation and prove the Rank-Nullity Theorem for linear transformations.

### LTR-0050: Image and Kernel of a Linear Transformation

#### The Image of a Linear Transformation

*image*of , denoted by , is the set In other words, the image of consists of individual images of all vectors of .

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,

So, 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 of VSP-0040, we know that

We can see that , so .

To identify vectors that span , we turn to Procedure proc:colspace of VSP-0040. 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

*kernel*of , denoted by , is the set In other words, the kernel of consists of all vectors of that map to in .

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 (Theorem th:span_is_subspace of VSP-0020). 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 (Theorem th:nullsubspacern of VSP-0040). 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 of VSP-0040), 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 get

But 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,