Existence of the Inverse of a Linear Transformation

In Exploration ep:inverse of Composition and Inverses of Linear Transformations we examined a linear transformation \(T:\RR ^2\rightarrow \RR ^2\) that doubles all input vectors, and its inverse \(S:\RR ^2\rightarrow \RR ^2\), that halves all input vectors. We observed that the composite functions \(S\circ T\) and \(T\circ S\) are both identity transformations. Diagrammatically, we can represent \(T\) and \(S\) as follows:

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This gives us a way of thinking about an inverse of \(T\) as a transformation that “undoes" the action of \(T\) 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 \(T:V\rightarrow W\), “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" \(T\) may fail to produce a function.

First, if two distinct vectors \(\vec {v}_1\) and \(\vec {v}_2\) map to the same vector \(\vec {w}\) in \(W\), then reversing the arrows gives us a mapping that is clearly not a function.

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Second, observe that our definition of an inverse of \(T:V\rightarrow W\) requires that the domain of the inverse transformation be \(W\). (Definition def:inverse, Composition and Inverses of Linear Transformations) If there is a vector \(\vec {b}\) in \(W\) that is not an image of any vector in \(V\), then \(\vec {b}\) cannot be in the domain of an inverse transformation.

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We now illustrate these potential issues with specific examples.

Our next goal is to develop vocabulary that would allow us to discuss issues illustrated in Figures \(1\) and \(2\).

One-to-one Linear Transformations

Figure \(1\) gave us a diagrammatic representation of a transformation that maps two distinct elements, \(\vec {v}_1\) and \(\vec {v}_2\) to the same element \(\vec {w}\), 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.

The transformation in Figure \(1\) is not one-to-one because \(\vec {v}_1\) and \(\vec {v}_2\) map to the same vector \(\vec {w}\), (i.e. \(T(\vec {v}_1)=T(\vec {v}_2)\)), yet the diagram suggests that \(\vec {v}_1\neq \vec {v}_2\).

Example 6 provides us with an important insight. Recall that the collection of all vectors that map to zero under a linear transformation constitute the kernel of the linear transformation. (See Image and Kernel of a Linear Transformation) It is clear that when \(\text {ker}(T)\neq \{\vec {0}\}\) the transformation is not one-to-one. This implication goes the other way as well. In Problem you will show that a linear transformation \(T\) is one-to-one if and only if \(\text {ker}(T)=\{\vec {0}\}\).

Since transformation in Example 3 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.

“Onto" Linear Transformations

Figure \(2\) 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.

Existence of Inverses

We will first assume that \(T\) is one-to-one and onto, and show that there exists a transformation \(S:W\rightarrow V\) such that \(S\circ T=\id _V\) and \(T\circ S=\id _W\). Because \(T\) is onto, for every \(\vec {w}\) in \(W\), there exists \(\vec {v}\) in \(V\) such that \(T(\vec {v})=\vec {w}\). Moreover, because \(T\) is one-to-one, vector \(\vec {v}\) is the only vector that maps to \(\vec {w}\). To stress this, we will say that for every \(\vec {w}\), there exists \(\vec {v}_{\vec {w}}\) such that \(T(\vec {v}_{\vec {w}})=\vec {w}\). (Since every \(\vec {v}\) maps to exactly one \(\vec {w}\), this notation makes sense for elements of \(V\) as well.) We can now define \(S:W\rightarrow V\) by \(S(\vec {w})=\vec {v}_{\vec {w}}\). Then

\[(S\circ T)(\vec {v}_{\vec {w}})=S(T(\vec {v}_{\vec {w}}))=S(\vec {w})=\vec {v}_{\vec {w}}\]
\[(T\circ S)(\vec {w})=T(S(\vec {w}))=T(\vec {v}_{\vec {w}})=\vec {w}\]
We conclude that \(S\circ T=\id _V\) and \(T\circ S=\id _W\). Therefore \(S\) is an inverse of \(T\).

We will now assume that \(T\) has an inverse \(S\) and show that \(T\) must be one-to-one and onto. Suppose

\[T(\vec {v}_1)=T(\vec {v}_2)\]
then
\[S(T(\vec {v}_1))=S(T(\vec {v}_2))\]
but then
\[\vec {v}_1=\vec {v}_2\]
We conclude that \(T\) is one-to-one.

Now suppose that \(\vec {w}\) is in \(W\). We need to show that some element of \(V\) maps to \(\vec {w}\). Let \(\vec {v}=S(\vec {w})\). Then

\[T(\vec {v})=T(S(\vec {w}))=(T\circ S)(\vec {w})=\id _W(\vec {w})=\vec {w}\]
We conclude that \(T\) is onto.

Uniqueness of Inverses

Definition ?? of Composition and Inverses of Linear Transformations refers to \(S\) as an inverse of \(T\), implying that there may be more than one such transformation \(S\). We will now show that if such a transformation \(S\) exists, it is unique. This will allow us to refer to it as the inverse of \(T\) and to start using \(T^{-1}\) to denote the unique inverse of \(T\).

Let \(T:V\rightarrow W\) be a linear transformation. If \(S\) is an inverse of \(T\), then \(S\) satisfies

\[S\circ T=\id _V\quad \text {and}\quad T\circ S=\id _W\]

Suppose there is another transformation, \(S'\), such that

\[S'\circ T=\id _V\quad \text {and}\quad T\circ S'=\id _W\]
We now show that \(S=S'\).
\[S=S\circ \id _W=S\circ (T\circ S')=(S\circ T)\circ S'=\id _V\circ S'=S'\]

Practice Problems

Show that a linear transformation \(T:\RR ^2\rightarrow \RR ^2\) with standard matrix \(A=\begin{bmatrix}2&-4\\-3&6\end{bmatrix}\) is not one-to-one.
Show that multiple vectors map to \(\vec {0}\).
Show that a linear transformation \(T:\RR ^2\rightarrow \RR ^3\) with standard matrix \(A=\begin{bmatrix}1&2\\-1&1\\0&1\end{bmatrix}\) is not onto.
Find \(\vec {b}\) such that \(A\vec {x}=\vec {b}\) has no solutions.
Suppose that a linear transformation \(T:\RR ^3\rightarrow \RR ^3\) has a standard matrix \(A\) such that \(\text {rref}(A)=I\).

Prove that \(T\) is one-to-one.

How many solutions does \(A\vec {x}=\vec {b}\) have?
Prove that \(T\) is onto.
Does \(A\vec {x}=\vec {b}\) have a solution for every \(\vec {b}\)?
Define a transformation \(T:\RR ^2\rightarrow \RR ^2\) by
\[T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}x\\-2x+4y\end{bmatrix}\]
Show that \(T\) is a linear transformation that has an inverse.
You will need to demonstrate that \(T\) is one-to-one and onto.
Let \(V=\text {span}\left (\begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\0\end{bmatrix}\right )\). Define a linear transformation \(T:V\rightarrow \RR ^2\) by
\[T\left (\begin{bmatrix}1\\0\\1\end{bmatrix}\right )=\begin{bmatrix}0\\1\end{bmatrix}\quad \text {and}\quad T\left (\begin{bmatrix}0\\1\\0\end{bmatrix}\right )=\begin{bmatrix}-1\\1\end{bmatrix}\]
Prove that \(T\) has an inverse.
Prove that a linear transformation \(T:V\rightarrow W\) is one-to-one if and only if \(\text {ker}(T)=\{\vec {0}\}\).