es

Subspaces

Homogeneous systems

Duality

Intersection

Direct sums

Image & Kernel

For any map f : E ↦ F between sets, and any subset X ⊂ V, its image is denoted as f(X) = { f(x) : x ∈ X} ⊂ Y. The image of f is
\[ \mbox{im}(f) = \left\{ f(x) \,:\ x \in X \right\} \subset F. \]
The inverse image of a subset Y ⊂ F is
\[ f^{-1} (Y) = \left\{ x \in E \, : \ f(x) \in Y \right\} \subset E . \]
Note that this definition does not require f to be invertible, but when it is, the inverse image of Y is also the image of Y by the inverse of f so that the notation is coherent.
Theorem 8: Let T : V ⇾ U be linear transformation.
  1. The inverse image T−1(Y) of any vector subspace Y ⊂ U is a vector subspace of V.
  2. The image T(X) of any vector subspace X ⊂ V is a subspace of U.
  1. Let us consider a subspace Y of V. If x, u ∈ T-1, namely, T(x), T(u) ∈ Y. Then for k ∈ 𝔽, we have \[ T(k{\bf x} + {\bf u}) = k\,T({\bf x}) + T({\bf u}) \in Y. \] This means that kx + u ∈ T-1(Y). This proves the first assertion.
  2. Let now X be any subspace of U. Take two elements of T(X). They can be written in the form T(x) and T(u) for some x, u ∈ X. Then \[ T(k{\bf x} + {\bf u}) = k\,T({\bf x}) + T({\bf u}) \in T(X) \] is in the image of X. Hence, T(X) is a subspace of V.
   
Example 10:    ■
End of Example 10
Corollary 2: For any linear transformation T : V ⇾ U, T−1(0U) is a subspace of V and im(T) = T(V) is a subspace of U.
Corollary 3: For any two linear transformations of 𝔽-vector space V into another 𝔽-vector space U, the coincidence subset \[ \left\{ {\bf x} \in V \ : \ f({\bf x}) = g({\bf x}) \right\} \] is a vector subspace of V.
The coincidence subset of f and g consists precisely of the x ∈ U such that \[ \left( f - g \right) ({\bf x}) = f({\bf x}) - g ({\bf x}) = {\bf 0}_V . \] So it is the inverse image of the subspace {0} by the linear map (f − g ).
   
Example 11:    ■
End of Example 11
Let T : V ⇾ U be a linear transformation of two vector spaces over the same scalar field. The kernel of T is the subspace ker(T) = T−1(0) of V. The image of T is the subspace T(V) of U. The nullity of T is the dimension (when finite) of its kernel. The rank of T is the dimension (when finite) of its image.
If a linear map T : V ⇾ V is injective (one-to-one), then 0 ∈ U is the only element having image 0 ∈ U, hence ker(T) = {0}. A converse statement holds.
Theorem 9: For a linear transformation T : U ⇾ V between two vector spaces, \[ \mbox{ker}(T) = \left\{ {\bf 0} \right\} \qquad \iff \qquad T \mbox{ is injective}. \]
For any linear mao \[ T({\bf x}) = T({\bf y}) \qquad \ff \qquad T({\bf x}) - T({\bf y}) = {\bf 0} \] If ker(T) = {0}, we have \[ T({\bf x}) = T({\bf y}) \qquad \iff \qquad {\bf x} - {\bf y} = {\bf 0} \qquad \iff \qquad {\bf x} = {\bf y} . \] This proves that T is injective.
   
Example 12:    ■
End of Example 12

 

   

 

  1. Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
  2. Beezer, R.A., A First Course in Linear Algebra, 2017.
  3. Fitzpatrick, S., Linear Algebra: A second course, featuring proofs and Python. 2023.