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.
Algebra / English
Rank and Nullity
IIT Roorkee July 2018 · YouTube · 33:31
YouTube · IIT Roorkee July 2018Original video
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Develop rank and nullity for linear maps, prove the dimension relation and use it to test injectivity and surjectivity.
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Chapters
3:00Kernel and range in an example19:30Extend a kernel basis25:30A dimension obstructionLearning script
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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.
Knowledge cards
Extend a kernel basis
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 dimension obstruction
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.
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- Rank and nullity ExplanationAt 3:00
Why this connection?
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.
Questions this video answers
Find a method↗
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.
YouTubeRank and Nullity
3:00 – 19:30Watch this moment ↗
Understand why↗
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.
YouTubeRank and Nullity
19:30 – 33:31Watch this moment ↗
Meet the concept↗
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.
YouTubeRank and Nullity
0:00 – 3:00Watch this moment ↗
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