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.
- The transformation A is invertible if and only if \mathcal{R}(A)=\mathcal{V} and \mathcal{N}(A)=\mathcal{O} .
- 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} .
- The subspace \mathcal{M} is invariant under A if and only if A\mathcal{M} \subset \mathcal{M} .
- 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} .
- 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.
- Give an example of a linear transformation on a three-dimensional space with a two-dimensional range.
- 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.
- Two projections E and F have the same range if and only if E F=F and F E=E
- 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.
