Kernel and range in an example
For T(x₁,x₂,x₃)=(x₁−x₂,2x₃), the kernel is spanned by (1,1,0) and the range is all of the two-dimensional codomain. The nullity is one and the rank is two.
IIT Roorkee July 2018 · YouTube · 33:31
Develop rank and nullity for linear maps, prove the dimension relation and use it to test injectivity and surjectivity.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
For T(x₁,x₂,x₃)=(x₁−x₂,2x₃), the kernel is spanned by (1,1,0) and the range is all of the two-dimensional codomain. The nullity is one and the rank is two.
Extend a basis of the kernel to a basis of the finite-dimensional domain. Images of the added basis vectors form a basis of the range, giving nullity+rank=domain dimension.
A linear map from a five-dimensional space to a three-dimensional space cannot be injective: its rank is at most three, so its nullity is at least two.
For T(x₁,x₂,x₃)=(x₁−x₂,2x₃), the kernel is spanned by (1,1,0) and the range is all of the two-dimensional codomain. The nullity is one and the rank is two.
Extend a basis of the kernel to a basis of the finite-dimensional domain. Images of the added basis vectors form a basis of the range, giving nullity+rank=domain dimension.
A linear map from a five-dimensional space to a three-dimensional space cannot be injective: its rank is at most three, so its nullity is at least two.
For T(x₁,x₂,x₃)=(x₁−x₂,2x₃), the kernel is spanned by (1,1,0) and the range is all of the two-dimensional codomain. The nullity is one and the rank is two. Extend a basis of the kernel to a basis of the finite-dimensional domain. Images of the added basis vectors form a basis of the range, giving nullity+rank=domain dimension. A linear map from a five-dimensional space to a three-dimensional space cannot be injective: its rank is at most three, so its nullity is at least two.
Extend a basis of the kernel to a basis of the finite-dimensional domain. The images of the added basis vectors form a basis of the range, giving nullity+rank=domain dimension.
Conditions: The linear map is defined on a finite-dimensional domain.; A basis for the kernel has been identified.
A linear map from a five-dimensional space to a three-dimensional space cannot be injective because its rank is at most three, so its nullity is at least two.
Conditions: The domain is five-dimensional.; The codomain is three-dimensional.; The map is linear.
The kernel is spanned by the vector (1,1,0), and the range is all of the two-dimensional codomain. The nullity is one and the rank is two.
Conditions: The linear map is defined as T(x₁,x₂,x₃)=(x₁−x₂,2x₃).; The domain is three-dimensional and the codomain is two-dimensional.