Definition 1. If \(A\) is a linear transformation on a vector space \(\mathcal{V}\) and if \(\mathcal{M}\) is a subspace of \(\mathcal{V}\) , the image of \(\mathcal{M}\) under \(A\) , in symbols \(A \mathcal{M}\) , is the set of all vectors of the form \(A x\) with \(x\) in \(\mathcal{M}\) . The range of \(A\) is the set \(\mathcal{R}(A)=A \mathcal{V}\) ; the null-space of \(A\) is the set \(\mathcal{N}(A)\) of all vectors \(x\) for which \(A x=0\) .

It is immediately verified that \(A\mathcal{M}\) and \(\mathcal{N}(A)\) are subspaces. If, as usual, we denote by \(\mathcal{O}\) the subspace containing the vector \(0\) only, it is easy to describe some familiar concepts in terms of the terminology just introduced; we list some of the results.

  1. The transformation \(A\) is invertible if and only if \(\mathcal{R}(A)=\mathcal{V}\) and \(\mathcal{N}(A)=\mathcal{O}\) .
  2. In case \(\mathcal{V}\) is finite-dimensional, \(A\) is invertible if and only if \(\mathcal{R}(A)=\mathcal{V}\) or \(\mathcal{N}(A) = \mathcal{O}\) .
  3. The subspace \(\mathcal{M}\) is invariant under \(A\) if and only if \(A\mathcal{M} \subset \mathcal{M}\) .
  4. A pair of complementary subspaces \(\mathcal{M}\) and \(\mathcal{N}\) reduce \(A\) if and only if \(A\mathcal{M} \subset \mathcal{M}\) and \(A\mathcal{N} \subset \mathcal{N}\) .
  5. If \(E\) is the projection on \(\mathcal{M}\) along \(\mathcal{N}\) , then \(\mathcal{R}(E) = \mathcal{M}\) and \(\mathcal{N}(E) = \mathcal{N}\) .

All these statements are easy to prove; we indicate the proof of (v). From Section: Projections , Theorem 2, we know that \(\mathcal{N}\) is the set of all solutions of the equation \(E x=0\) ; this coincides with our definition of \(\mathcal{N}(E)\) . We know also that \(\mathcal{M}\) is the set of all solutions of the equation \(E x=x\) . If \(x\) is in \(\mathcal{M}\) , then \(x\) is also in \(\mathcal{R}(E)\) , since \(x\) is the image under \(E\) of something (namely of \(x\) itself). Conversely, if a vector \(x\) is the image under \(E\) of something, say, \(x=E y\) (so that \(x\) is in \(\mathcal{R}(E)\) ), then \(E x=E^{2} x=E y=x\) , so that \(x\) is in \(\mathcal{M}\) .

Warning: it is accidental that for projections \(\mathcal{R} \oplus \mathcal{N} = \mathcal{V}\) . In general it need not even be true that \(\mathcal{R} = \mathcal{R}(A)\) and \(\mathcal{N} = \mathcal{N}(A)\) are disjoint. It can happen, for example, that for a certain vector \(x\) we have \(x \neq 0\) , \(A x \neq 0\) , and \(A^{2} x=0\) ; for such a vector, \(A x\) clearly belongs to both the range and the null-space of \(A\) .

Theorem 1. If \(A\) is a linear transformation on a vector space \(\mathcal{V}\) , then \[(\mathcal{R}(A))^{0}=\mathcal{N}(A^{\prime}); \tag{1}\] if \(\mathcal{V}\) is finite-dimensional, then \[(\mathcal{N}(A))^{0}=\mathcal{R}(A^{\prime}). \tag{2}\] 

Proof. If \(y\) is in \((\mathcal{R}(A))^{0}\) , then, for all \(x\) in \(\mathcal{V}\) , \[0=[A x, y]=[x, A^{\prime} y],\] so that \(A^{\prime} y=0\) and \(y\) is in \(\mathcal{N}(A^{\prime})\) . If, on the other hand, \(y\) is in \(\mathcal{N}(A^{\prime})\) , then, for all \(x\) in \(\mathcal{V}\) , \[0=[x, A^{\prime} y]=[A x, y],\] so that \(y\) is in \((\mathcal{R}(A))^{0}\) .

If we apply (1) to \(A^{\prime}\) in place of \(A\) , we obtain \[(\mathcal{R}(A^{\prime}))^{0}=\mathcal{R}(A^{\prime \prime}) . \tag{3}\] If \(\mathcal{V}\) is finite-dimensional (and hence reflexive), we may replace \(A^{\prime \prime}\) by \(A\) in (3), and then we may form the annihilator of both sides; the desired conclusion (2) follows from Section: Annihilators , Theorem 2. ◻

EXERCISES

Exercise 1. Use the differentiation operator on \(\mathcal{P}_{n}\) to show that the range and the null-space of a linear transformation need not be disjoint.

Exercise 2. 

  1. Give an example of a linear transformation on a three-dimensional space with a two-dimensional range.
  2. Give an example of a linear transformation on a three-dimensional space with a two-dimensional null-space.

Exercise 3. Find a four-by-four matrix whose range is spanned by \((1,0,1,0)\) and \((0,1,0,1)\) .

Exercise 4. 

  1. Two projections \(E\) and \(F\) have the same range if and only if \(E F=F\) and \(F E=E\) 
  2. Two projections \(E\) and \(F\) have the same null-space if and only if \(E F=E\) and \(F E=F\) .

Exercise 5. If \(E_{1}, \ldots, E_{k}\) are projections with the same range and if \(\alpha_{1}, \ldots, \alpha_{k}\) are scalars such that \(\sum_{i} \alpha_{i}=1\) , then \(\sum_{i} \alpha_{i} E_{i}\) is a projection.