A linear map (or transformation, or function) transforms elements of a vector space called domain into elements of another vector space called codomain.

The kernel (or null space) of a linear transformation is the subset of the domain that is transformed into the zero vector.

Let us provide a more formal definition of kernel.

Definition Let and be two vector spaces. Let be a linear map. The setis called the null space (or kernel) of .

Let us see some examples.

Example Let be the space of all column vectors having real entries. Let be the linear map defined by the matrix productwhere For any , denote by and the two entries of , so thatAs a consequence, whenever . Therefore, the kernel of is formed by all the vectors of whose two entries are equal to each other:

Example Let and respectively be the spaces of all and column vectors having real entries. Let be the linear map defined by the matrix productwhere For any , denote by and the two entries of , so thatBy looking at the three entries of , it is apparent that only when . Therefore the kernel of is formed by a single vector, the zero vector:

An interesting property of the null space is that it is a subspace of the domain, that is, it is closed with respect to taking linear combinations.

Proposition Let and be two vector spaces. Let be a linear map. Then, the null space is a subspace of .

Proof

By the definition of subspace, is a subspace of if and only if any linear combination of elements of belongs to . Let us check that this condition is verified. Arbitrarily choose two vectors and two scalars and . Then,where: in step we have used the fact that is a linear map; in step we have used the fact that and, as a consequence, . Thus, the linear combinationbelongs to the kernel (because the function maps it into the zero vector). This is exactly what we needed to prove.

Note that the zero vector always belongs to the kernel. In fact, the linearity of implies thatfor any and any scalar . Thus, when we set , the previous equation becomes

Below you can find some exercises with explained solutions.

Let be the space of all column vectors having real entries. Let be the linear map defined by where

Find the null space of .

Solution

For any , denote by and the two entries of , so thatHence, for and any value of . Thus, the null space of is:

Remember that a linear transformation can be defined by specifying the values taken by in correspondence of a basis of (see the lecture on linear maps).

Let be a basis for . Let be a basis for . Suppose that is defined byFind the kernel of .

Solution

Any vector can be represented in terms of the basis aswhere are scalars. Then, by the linearity of , we have thatSo, we have that wheneverThus,

Let and respectively be the spaces of all and column vectors having real entries. Let be the linear map defined by the matrix productwhere is a matrix.

Find the null space of under the hypothesis that the columns of are linearly independent.

Solution

Denote by the three columns of . For any , denote by the three entries of . The product can be written as a linear combination of the columns of with coefficients taken from the vector :Since the three columns of are linearly independent, the only linear combination that givesis the combination with coefficients . Therefore,

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