Answer to Question #167050 in Linear Algebra for Sarita bartwal

Question

Given any n∈N , it is possible to define a linear transformation whose kernel has dimension n.

True or false with full explanation

Answer 1

Let T be a linear map from $R_ 5$  to $R_ 3$ . What are the possible dimensions of the kernel of The kernel is the set of vectors in the domain that are mapped to zero in the codomain. The dimension of the kernel can be thought of as the number of dimensions that get ‘squashed’ by the transformation.

By ‘squashed’, I mean, for example, all of the vectors in a 3

3-dimensional space being mapped to a 2

2-dimensional plane. You can imagine a cube, or some other 3

3-dimensional object, being squashed until it is flat

We are mapping from a 5-dimensional space to a 3-dimensional space, so we are already forced to squash 2 dimensions. Therefore the dimension of the kernel is at least 2. If all of the vectors are mapped to zero by the transformation, then all 5

5 dimensions of the domain will be squashed, meaning that the dimension of the kernel is at most 5

5. So we have $2≤dim(Null(T))≤5,2≤dim(Null(T))≤5.$

If you want to use the rank-nullity theorem, we can instead consider the image of T. In the case where all vectors are mapped to zero, the image clearly has dimension zero. It is also clear that the dimension of the image can be at most 3, which will be the case if the ‘output’ vectors occupy all of the space we are mapping to. So we have $0≤dim(Im(T))≤3$  which, by the rank-nullity theorem $(dim(Im(T))+dim(Null(T))=5$ in this case), implies the result above.