值域与零空间

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

It is immediately verified that $A ℳ$ and $𝒩 (A)$ are subspaces. If, as usual, we denote by $𝒪$ 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 $ℛ (A) = 𝒱$ and $𝒩 (A) = 𝒪$ .
In case $𝒱$ is finite-dimensional, $A$ is invertible if and only if $ℛ (A) = 𝒱$ or $𝒩 (A) = 𝒪$ .
The subspace $ℳ$ is invariant under $A$ if and only if $A ℳ \subset ℳ$ .
A pair of complementary subspaces $ℳ$ and $𝒩$ reduce $A$ if and only if $A ℳ \subset ℳ$ and $A 𝒩 \subset 𝒩$ .
If $E$ is the projection on $ℳ$ along $𝒩$ , then $ℛ (E) = ℳ$ and $𝒩 (E) = 𝒩$ .

All these statements are easy to prove; we indicate the proof of (v). From Section: Projections , Theorem 2, we know that $𝒩$ is the set of all solutions of the equation $E x = 0$ ; this coincides with our definition of $𝒩 (E)$ . We know also that $ℳ$ is the set of all solutions of the equation $E x = x$ . If $x$ is in $ℳ$ , then $x$ is also in $ℛ (E)$ , 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 $ℛ (E)$ ), then $E x = E^{2} x = E y = x$ , so that $x$ is in $ℳ$ .

Warning: it is accidental that for projections $ℛ \oplus 𝒩 = 𝒱$ . In general it need not even be true that $ℛ = ℛ (A)$ and $𝒩 = 𝒩 (A)$ 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 $𝒱$ , then if $𝒱$ is finite-dimensional, then

Proof. If $y$ is in $(ℛ (A))^{0}$ , then, for all $x$ in $𝒱$ , so that and $y$ is in . If, on the other hand, $y$ is in , then, for all $x$ in $𝒱$ , so that $y$ is in $(ℛ (A))^{0}$ .

If we apply (1) to in place of $A$ , we obtain If $𝒱$ is finite-dimensional (and hence reflexive), we may replace 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 $𝒫_{n}$ 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 $E_{1}, \dots, E_{k}$ are projections with the same range and if $α_{1}, \dots, α_{k}$ are scalars such that $\sum_{i} α_{i} = 1$ , then $\sum_{i} α_{i} E_{i}$ is a projection.