To every linear transformation we associate two important subspaces.
let be linear.
the kernel of , written , is
the image of , written , is
In other words, the image is what we normally mean by the image of a function.
An important family of examples are the linear maps defined by left-multiplication by an matrix with entries from the field . In that case the image is equal to the column space by Proposition 3.2.1, and the kernel is the nullspace .
Let be a linear map. Then .
by the first part of the definition of linearity. Now add to both sides:
Let be linear. Then and .
To show something is a subspace you must check the three conditions: it contains the zero vector, it is closed under addition, it is closed under scalar multiplication.
First, the kernel.
To show that the kernel contains , we must show that . That’s exactly Lemma 4.15.1.
If then , so .
If and then by the second part of the definition of linearity, and this is which equals . Since , we have .
Next, the image.
We know from Lemma 4.15.1 that , so .
Any two elements of have the form some . Then (linearity definition part 1), which is an element if , so is closed under addition.
If and then by the definition of linearity part 2, and this is an element of as it is applied to something, so is closed under scalar multiplication.
∎
Let so that we have a linear map given by . We will find and .
Another way to write this is that , and so .
Now we’ll do the kernel.
Again we could write this as . The kernel and image are equal in this case.
Let be . We will describe and .
A polynomial has derivative zero if and only if it is constant, so is the set of all constant polynomials. This is spanned by any (nonzero) constant polynomial, so it has dimension one.
Next consider . Let be the subspace spanned by , that is, the subspace consisting of all polynomials of degree at most . Certainly , since when you differentiate a polynomial of degree at most n you get a polynomial of degree at most . But if then has an indefinite integral in and , so every is in , so .