5.5: One-to-One and Onto Transformations
- Determine if a linear transformation is onto or one to one.
Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. We define the range or image of \(T\) as the set of vectors of \(\mathbb{R}^{m}\) which are of the form \(T \left(\vec{x}\right)\) (equivalently, \(A\vec{x}\)) for some \(\vec{x}\in \mathbb{R}^{n}\). It is common to write \(T\mathbb{R}^{n}\), \(T\left( \mathbb{R}^{n}\right)\), or \(\mathrm{Im}\left( T\right)\) to denote these vectors.
Let \(A\) be an \(m\times n\) matrix where \(A_{1},\cdots , A_{n}\) denote the columns of \(A.\) Then, for a vector \(\vec{x}=\left [ \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right ]\) in \(\mathbb{R}^n\),
\[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]
Therefore, \(A \left( \mathbb{R}^n \right)\) is the collection of all linear combinations of these products.
- Proof
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This follows from the definition of matrix multiplication.
Below is a video on determining if function mappings are one to one and/or onto.
This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. We define them now.
Suppose \(\vec{x}_1\) and \(\vec{x}_2\) are vectors in \(\mathbb{R}^n\). A linear transformation \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) is called one to one (often written as \(1-1)\) if whenever \(\vec{x}_1 \neq \vec{x}_2\) it follows that : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber \]
Equivalently, if \(T\left( \vec{x}_1 \right) =T\left( \vec{x}_2\right) ,\) then \(\vec{x}_1 = \vec{x}_2\). Thus, \(T\) is one to one if it never takes two different vectors to the same vector.
Below is a video on one to one transformations.
The second important characterization is called onto.
Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Then \(T\) is called onto if whenever \(\vec{x}_2 \in \mathbb{R}^{m}\) there exists \(\vec{x}_1 \in \mathbb{R}^{n}\) such that \(T\left( \vec{x}_1\right) = \vec{x}_2.\)
We often call a linear transformation which is one-to-one an injection . Similarly, a linear transformation which is onto is often called a surjection .
Below is a video on one to one and onto transformation definitions.
Below is a video on onto transformations.
The following proposition is an important result.
Let \(T:\mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\).
- Proof
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We need to prove two things here. First, we will prove that if \(T\) is one to one, then \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x}=\vec{0}\). Second, we will show that if \(T(\vec{x})=\vec{0}\) implies that \(\vec{x}=\vec{0}\), then it follows that \(T\) is one to one. Recall that a linear transformation has the property that \(T(\vec{0}) = \vec{0}\).
Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). \[T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0})\nonumber \] and so, adding the additive inverse of \(T(\vec{0})\) to both sides, one sees that \(T(\vec{0})=\vec{0}\). If \(T(\vec{x})=\vec{0}\) it must be the case that \(\vec{x}=\vec{0}\) because it was just shown that \(T(\vec{0})=\vec{0}\) and \(T\) is assumed to be one to one.
Now assume that if \(T(\vec{x})=\vec{0},\) then it follows that \(\vec{x}=\vec{0}.\) If \(T(\vec{v})=T(\vec{u}),\) then \[T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0}\nonumber \] which shows that \(\vec{v}-\vec{u}=0\). In other words, \(\vec{v}=\vec{u}\), and \(T\) is one to one.
Note that this proposition says that if \(A=\left [ \begin{array}{ccc} A_{1} & \cdots & A_{n} \end{array} \right ]\) then \(A\) is one to one if and only if whenever \[0 = \sum_{k=1}^{n}c_{k}A_{k}\nonumber \] it follows that each scalar \(c_{k}=0\).
Below is a video on comparing one to one and onto matrix transformations.
Below is a video on determining if transformation mappings are one to one and/or onto.
We will now take a look at an example of a one to one and onto linear transformation.
Suppose \[T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{r} x \\ y \end{array} \right ]\nonumber \] Then, \(T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}\) is a linear transformation. Is \(T\) onto? Is it one to one?
Solution
Recall that because \(T\) can be expressed as matrix multiplication, we know that \(T\) is a linear transformation. We will start by looking at onto. So suppose \(\left [ \begin{array}{c} a \\ b \end{array} \right ] \in \mathbb{R}^{2}.\) Does there exist \(\left [ \begin{array}{c} x \\ y \end{array} \right ] \in \mathbb{R}^2\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ] ?\) If so, then since \(\left [ \begin{array}{c} a \\ b \end{array} \right ]\) is an arbitrary vector in \(\mathbb{R}^{2},\) it will follow that \(T\) is onto.
This question is familiar to you. It is asking whether there is a solution to the equation \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber \] This is the same thing as asking for a solution to the following system of equations. \[\begin{array}{c} x+y=a \\ x+2y=b \end{array}\nonumber \] Set up the augmented matrix and row reduce. \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. Therefore, we have shown that for any \(a, b\), there is a \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). Thus \(T\) is onto.
Now we want to know if \(T\) is one to one. By Proposition \(\PageIndex{1}\) it is enough to show that \(A\vec{x}=0\) implies \(\vec{x}=0\). Consider the system \(A\vec{x}=0\) given by: \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \]
This is the same as the system given by
\[\begin{array}{c} x + y = 0 \\ x + 2y = 0 \end{array}\nonumber \]
We need to show that the solution to this system is \(x = 0\) and \(y = 0\). By setting up the augmented matrix and row reducing, we end up with \[\left [ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right ]\nonumber \]
This tells us that \(x = 0\) and \(y = 0\). Returning to the original system, this says that if
\[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \]
then \[\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \]
In other words, \(A\vec{x}=0\) implies that \(\vec{x}=0\). By Proposition \(\PageIndex{1}\) , \(A\) is one to one, and so \(T\) is also one to one.
We also could have seen that \(T\) is one to one from our above solution for onto. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). Therefore, there is only one vector, specifically \(\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 2a-b\\ b-a \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). Hence by Definition \(\PageIndex{1}\) , \(T\) is one to one.
Let \(T: \mathbb{R}^4 \mapsto \mathbb{R}^2\) be a linear transformation defined by \[T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber \] Prove that \(T\) is onto but not one to one.
Solution
You can prove that \(T\) is in fact linear.
To show that \(T\) is onto, let \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) be an arbitrary vector in \(\mathbb{R}^2\). Taking the vector \(\left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] \in \mathbb{R}^4\) we have \[T \left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} x + 0 \\ y + 0 \end{array} \right ] = \left [ \begin{array}{c} x \\ y \end{array} \right ]\nonumber \] This shows that \(T\) is onto.
By Proposition \(\PageIndex{1}\) \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). Observe that \[T \left [ \begin{array}{r} 1 \\ 0 \\ 0 \\ -1 \end{array} \right ] = \left [ \begin{array}{c} 1 + -1 \\ 0 + 0 \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \] There exists a nonzero vector \(\vec{x}\) in \(\mathbb{R}^4\) such that \(T(\vec{x}) = \vec{0}\). It follows that \(T\) is not one to one.
Below is a video on determining if a linear transformation is one to one and/or onto.
Below is another video on determining if a linear transformation is one to one and/or onto.
The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. It turns out that the matrix \(A\) of \(T\) can provide this information.
Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation induced by the \(m \times n\) matrix \(A\). Then \(T\) is one to one if and only if the rank of \(A\) is \(n\). \(T\) is onto if and only if the rank of \(A\) is \(m\).
Consider Example \(\PageIndex{2}\) . Above we showed that \(T\) was onto but not one to one. We can now use this theorem to determine this fact about \(T\).
Let \(T: \mathbb{R}^4 \mapsto \mathbb{R}^2\) be a linear transformation defined by \[T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber \] Prove that \(T\) is onto but not one to one.
Solution
Using Theorem \(\PageIndex{1}\) we can show that \(T\) is onto but not one to one from the matrix of \(T\). Recall that to find the matrix \(A\) of \(T\), we apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). The result is the \(2 \times 4\) matrix A given by \[A = \left [ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array} \right ]\nonumber \] Fortunately, this matrix is already in reduced row-echelon form. The rank of \(A\) is \(2\). Therefore by the above theorem \(T\) is onto but not one to one.
Below is a video on classifying matrix multiplication: onto and one to one.
Below is a video on determining if a 4x4 matrix represents a one to one and/or onto transformation.
Recall that if \(S\) and \(T\) are linear transformations, we can discuss their composite denoted \(S \circ T\). The following examines what happens if both \(S\) and \(T\) are onto.
Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. If \(T\) and \(S\) are onto, then \(S \circ T\) is onto.
Solution
Let \(\vec{z}\in \mathbb{R}^m\). Since \(S\) is onto, there exists a vector \(\vec{y}\in \mathbb{R}^n\) such that \(S(\vec{y})=\vec{z}\). Furthermore, since \(T\) is onto, there exists a vector \(\vec{x}\in \mathbb{R}^k\) such that \(T(\vec{x})=\vec{y}\). Thus \[\vec{z} = S(\vec{y}) = S(T(\vec{x})) = (ST)(\vec{x}),\nonumber \] showing that for each \(\vec{z}\in \mathbb{R}^m\) there exists and \(\vec{x}\in \mathbb{R}^k\) such that \((ST)(\vec{x})=\vec{z}\). Therefore, \(S \circ T\) is onto.
The next example shows the same concept with regards to one-to-one transformations.
Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Prove that if \(T\) and \(S\) are one to one, then \(S \circ T\) is one-to-one.
Solution
To prove that \(S \circ T\) is one to one, we need to show that if \(S(T (\vec{v})) = \vec{0}\) it follows that \(\vec{v} = \vec{0}\). Suppose that \(S(T (\vec{v})) = \vec{0}\). Since \(S\) is one to one, it follows that \(T (\vec{v}) = \vec{0}\). Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). Hence \(S \circ T\) is one to one.