We prove that a linear transformation has an inverse if and only if the transformation is “one-to-one” and “onto”.
In Exploration ep:inverse of LTR-0030 we examined a linear transformation that doubles all input vectors, and its inverse , that halves all input vectors. We observed that the composite functions and are both identity transformations. Diagrammatically, we can represent and as follows:
This gives us a way of thinking about an inverse of as a transformation that “undoes” the action of by “reversing” the mapping arrows. We will now use these intuitive ideas to understand which linear transformations are invertible and which are not.
Given an arbitrary linear transformation , “reversing the arrows” may not always result in a transformation. Recall that transformations are functions. The figures below show two ways in which our attempt to “reverse” may fail to produce a function.
First, if two distinct vectors and map to the same vector in , then reversing the arrows gives us a mapping that is clearly not a function.
Second, observe that our definition of an inverse of requires that the domain of the inverse transformation be . (Definition def:inverse, LTR-0030) If there is a vector in that is not an image of any vector in , then cannot be in the domain of an inverse transformation.
We now illustrate these potential issues with specific examples.
We now dig a little deeper to get additional insights into why does not have an inverse. Observe that all vectors of the form map to . To verify this, use matrix multiplication: This shows that there are infinitely many vectors that map to . So, “reversing the arrows” would not result in a function. (See Figure 1)
We now get another insight into why is not invertible. To find a vector such that no vector of maps to , we need to find for which the matrix equation
has no solution.
Let . Gauss-Jordan elimination yields:
Equation (ex:matrix) has a solution if and only if . Since we do not want (ex:matrix) to have a solution, all we need to do is pick values , and such that . Let . Then no element of maps to . This shows that we cannot “reverse the arrows” in an attempt to produce an inverse of . (See Figure 2)
Figure gave us a diagrammatic representation of a transformation that maps two distinct elements, and to the same element , making it impossible for us to “reverse the arrows” in an attempt to find the inverse transformation. Based on this example, it is reasonable to conjecture that for a transformation to be invertible, the transformation must be such that each output is the image of exactly one input. Such transformations are called one-to-one.
- Suppose Then It is clear that and are linearly independent. Therefore, we must have and . But then and , so
Since transformation in Example ex:notonto is one-to-one but not invertible we can conjecture that being one-to-one is a necessary, but not a sufficient condition for a linear transformation to have an inverse. We will consider the other necessary condition next.
Figure makes a convincing case that for a transformation to be invertible every element of the codomain must have something mapping to it. Transformations such that every element of the codomain is an image of some element of the domain are called onto.
- Let be an element of the codomain (). We need to find in the domain () such that . Observe that is invertible, and Let , then
Let be an element of . We need to show that there exists in such that . Observe that This means that has a solution (in fact, it has infinitely many solutions) for every in . Therefore every in is an image of some in . We conclude that is onto.
- We will now show that is one-to-one. Suppose
for some and in . Vectors and are in the span of and , so
for some scalars .
Thus, This implies that which, in turn, implies . This gives us , and we conclude that is one-to-one.
Next we will show that is onto. The key observation is that vectors and span . This means that given a vector in , we can write as . But this means that We conclude that is onto.
- We will first assume that is one-to-one and onto, and show that there
exists a transformation such that and . Because is onto, for every in , there
exists in such that . Moreover, because is one-to-one, vector is the only vector
that maps to . To stress this, we will say that for every , there exists such that
. (Since every maps to exactly one , this notation makes sense for elements of
as well.) We can now define by . Then
We conclude that and . Therefore is an inverse of .
We will now assume that has an inverse and show that must be one-to-one and onto. Suppose then but then We conclude that is one-to-one.
Now suppose that is in . We need to show that some element of maps to . Let . Then We conclude that is onto.
Recall that was introduced in Exploration init:subtosub of LTR-0030 to demonstrate that Theorem th:existunique of LTR-0030 is not always directly applicable. We now have additional tools. Theorem th:isomeansinvert assures us that has an inverse, but does not help us find it. We will visit this problem again, in LTR-0080, and find an inverse of .
Definition def:inverse of LTR-0030 refers to as an inverse of , implying that there may be more than one such transformation . We will now show that if such a transformation exists, it is unique. This will allow us to refer to the inverse of and to start using to denote the unique inverse of .
- Let be a linear transformation. If is an inverse of , then satisfies
Suppose there is another transformation, , such that We now show that .
Prove that is one-to-one.