References: edX online course MIT 8.05.1x Week 3.
Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 3.
We’ve looked at some basic properties of linear operators, so we’ll carry on with a few more definitions and theorems.
Null space and injectivity
First, we define the null space or kernel of an operator to be the set of all vectors which maps to the zero vector:
Theorem 1 The null space is a subspace.
Proof: Since is a linear operator, it maps the zero vector 0 to the zero vector, so . If two other vectors , then so is their sum, by the additivity property of linear operators. By the homogeneity property, if and , then . Thus is closed under addition and scalar multiplication, and contains the 0 vector, so it is a subspace.
An operator is injective if . That is, no two vectors are mapped to the same vector by the operator. An injective operator is also called one-to-one (or, in Zwiebach’s notes, two-to-two).
A useful result is the following:
Theorem 2 A linear operator is injective if and only if . That is, if the only vector mapped to 0 is 0 itself, then is one-to-one.
Proof: If is injective, then the only vector mapped to 0 is 0. To prove the converse, suppose and assume there are two different vectors such that . Then . However, since the only vector in the null space is 0, so we must have or . Thus is injective.
Range and surjectivity
The range of an operator is the set of all vectors produced by operating on vectors with . That is
The notation means the operator operates on vector space and has range .
Theorem 3 The range is a subspace.
Proof: As before, we must have , so the zero vector is in the range. To show that the range is closed under addition, choose two vectors . The corresponding vectors in the range are and . By linearity we must have , so is closed under addition. Similarly for scalar multiplication, if then .
An operator is surjective if , that is, if the range is the same as the original vector space upon which operates.
The null space and range of an operator obey the fundamental theorem of linear maps:
Theorem 4 If is finite-dimensional and then is finite-dimensional and
That is, the dimension of the null space and the dimension of the range add up to the dimension of the original vector space on which operates. Note that this makes sense only for finite-dimensional vector spaces.
Proof: Let be a basis of . If , we can add some more vectors to the basis to get a basis for (we haven’t proved this, but the proof is in Axler section 2.33). Suppose we need another vectors to do this. We then have a basis for :
The dimension of is therefore . To complete the proof, we need to show that , which we can do if we can show that is a basis for .
Since 4 is a basis for , we can write any vector as a linear combination
Operating on this equation with and using the fact that since the are a basis for the null space of , we have
Thus any vector in can be written as a linear combination of the vectors , so the vectors span the range of .
To complete the proof, we need to show that the vectors are linearly independent and thus form a basis for . To do this, suppose we have the equation
By linearity, we have
so and we can write
However, because the list 4 of vectors is a basis for , all the vectors in this list are linearly independent, which means that the only way two different linear combinations of these vectors can be equal is if all the coefficients are zero, that is, all the s and s are zero. Going back to 7, this means that the only solution is for all , which means that the vectors are linearly independent and span , so they form a basis for . Thus .