For $T:V\to W$ with $\dim V=n<\infty$, the following are equivalent: injectivity; $\ker T=\{0\}$; nullity zero; rank $n$; independent matrix columns; and the image of a domain basis being a basis of the range. Surjectivity onto the entire codomain follows only if its dimension also equals $n$.
Parents
Children
- Definition of one-to-one/injective linear transformationDefinition: A linear transformation \(T: V \to W\) is one-to-one (or injective ) if distinct vectors in the domain...
- Equivalence theorem for injective linear transformations: The inverse of T is a linear transformation on its range.A linear map is injective iff it has an inverse as a map $T:V\to T(V)$. That inverse is linear on the range. A global...
- Equivalence theorem for injective linear transformations: The null space of T is 0.A linear map is injective iff its null space is $\{0\}$. If $T(u)=T(v)$, linearity gives $T(u-v)=0$; a zero kernel...
- Equivalence theorem for injective linear transformations: The nullity of T is 0.For a linear map with finite-dimensional domain, injectivity is equivalent to nullity zero. Nullity is the dimension...
- Equivalence theorem for injective linear transformations: The kernel of T is 0.The kernel-zero criterion says $T$ is injective iff $\ker T=\{0 V\}$. The condition concerns the domain zero vector....
- Equivalence theorem for injective linear transformations: T(x)=0 only for x=0.For linear $T$, the statement “$T(x)=0$ only when $x=0$” is equivalent to injectivity. Compare two outputs:...
- Equivalence theorem for injective linear transformations: The rank of T is n.If $\dim V=n<\infty$, $T:V\to W$ is injective iff $\operatorname{rank}T=n$. Rank–nullity converts this to nullity...
- Equivalence theorem for injective linear transformations: The rank of T is equals the number of columns in any matrix representation..The columns of a matrix representation are independent iff its rank equals its column count, which is the domain...
- Equivalence theorem for injective linear transformations: The image of a basis for V is a basis for the range of T.The image of a domain basis always spans the range. It is also independent exactly when $T$ is injective. Thus $T$ is...
- Equivalence theorem for injective linear transformations: The columns of the matrix of T are linearly independent.With fixed bases, $T(v)=0$ becomes $A[v] B=0$. Thus $T$ is injective iff $A$ has only the zero null vector, iff its...
No children match.
Like how this is organised?
Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.