Can a Matrix Be One to One and Onto
5.5: One-to-Ane and Onto Transformations
- Page ID
- 14528
- Determine if a linear transformation is onto or 1 to i.
Allow \(T: \mathbb{R}^due north \mapsto \mathbb{R}^m\) be a linear transformation. Nosotros ascertain the range or image of \(T\) as the set of vectors of \(\mathbb{R}^{g}\) which are of the form \(T \left(\vec{x}\correct)\) (equivalently, \(A\vec{x}\)) for some \(\vec{x}\in \mathbb{R}^{due north}\). 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 \(A\) exist an \(m\times n\) matrix where \(A_{1},\cdots , A_{due north}\) denote the columns of \(A.\) So, for a vector \(\vec{x}=\left [ \begin{assortment}{c} x_{1} \\ \vdots \\ x_{north} \end{assortment} \correct ]\) in \(\mathbb{R}^n\),
\[A\vec{10}=\sum_{yard=1}^{n}x_{k}A_{one thousand}\nonumber \]
Therefore, \(A \left( \mathbb{R}^due north \right)\) is the collection of all linear combinations of these products.
- Proof
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This follows from the definition of matrix multiplication.
This department is devoted to studying two important characterizations of linear transformations, called ane to one and onto. We ascertain them now.
Suppose \(\vec{x}_1\) and \(\vec{x}_2\) are vectors in \(\mathbb{R}^n\). A linear transformation \(T: \mathbb{R}^n \mapsto \mathbb{R}^thou\) is called ane to one (frequently written as \(1-1)\) if whenever \(\vec{10}_1 \neq \vec{x}_2\) it follows that : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{ten}_2\right)\nonumber \]
Equivalently, if \(T\left( \vec{10}_1 \right) =T\left( \vec{10}_2\right) ,\) then \(\vec{x}_1 = \vec{ten}_2\). Thus, \(T\) is one to one if it never takes two different vectors to the same vector.
The 2nd important characterization is called onto.
Let \(T: \mathbb{R}^due north \mapsto \mathbb{R}^thou\) exist a linear transformation. And then \(T\) is called onto if whenever \(\vec{ten}_2 \in \mathbb{R}^{thousand}\) 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 i-to-one an injection. Similarly, a linear transformation which is onto is often chosen a surjection.
The following proposition is an important result.
Let \(T:\mathbb{R}^northward \mapsto \mathbb{R}^m\) be a linear transformation. Then \(T\) is one to one if and only if \(T(\vec{10}) = \vec{0}\) implies \(\vec{x}=\vec{0}\).
- Proof
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We need to testify 2 things hither. First, we will prove that if \(T\) is i to one, then \(T(\vec{10}) = \vec{0}\) implies that \(\vec{10}=\vec{0}\). Second, nosotros volition bear witness that if \(T(\vec{x})=\vec{0}\) implies that \(\vec{10}=\vec{0}\), then it follows that \(T\) is one to one. Call up that a linear transformation has the belongings that \(T(\vec{0}) = \vec{0}\).
Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). \[T(\vec{0})=T\left( \vec{0}+\vec{0}\correct) =T(\vec{0})+T(\vec{0})\nonumber \] and then, adding the additive inverse of \(T(\vec{0})\) to both sides, ane 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 presume that if \(T(\vec{x})=\vec{0},\) so it follows that \(\vec{x}=\vec{0}.\) If \(T(\vec{v})=T(\vec{u}),\) then \[T(\vec{v})-T(\vec{u})=T\left( \vec{five}-\vec{u}\correct) =\vec{0}\nonumber \] which shows that \(\vec{v}-\vec{u}=0\). In other words, \(\vec{5}=\vec{u}\), and \(T\) is one to one.
Note that this suggestion says that if \(A=\left [ \begin{assortment}{ccc} A_{1} & \cdots & A_{north} \end{array} \right ]\) so \(A\) is one to 1 if and only if whenever \[0 = \sum_{k=1}^{n}c_{g}A_{k}\nonumber \] information technology follows that each scalar \(c_{1000}=0\).
We volition now take a look at an instance of a one to one and onto linear transformation.
Suppose \[T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{rr} ane & 1 \\ 1 & two \end{array} \correct ] \left [ \begin{array}{r} x \\ y \end{array} \right ]\nonumber \] So, \(T:\mathbb{R}^{ii}\rightarrow \mathbb{R}^{2}\) is a linear transformation. Is \(T\) onto? Is it one to one?
Solution
Retrieve that because \(T\) can exist expressed as matrix multiplication, we know that \(T\) is a linear transformation. We will first by looking at onto. So suppose \(\left [ \brainstorm{array}{c} a \\ b \stop{assortment} \correct ] \in \mathbb{R}^{2}.\) Does in that location exist \(\left [ \begin{assortment}{c} x \\ y \end{array} \right ] \in \mathbb{R}^2\) such that \(T\left [ \brainstorm{array}{c} x \\ y \terminate{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \correct ] ?\) If so, then since \(\left [ \begin{array}{c} a \\ b \cease{array} \right ]\) is an arbitrary vector in \(\mathbb{R}^{ii},\) it will follow that \(T\) is onto.
This question is familiar to you. It is asking whether there is a solution to the equation \[\left [ \begin{array}{cc} one & 1 \\ ane & 2 \end{array} \correct ] \left [ \brainstorm{assortment}{c} ten \\ y \end{assortment} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \correct ]\nonumber \] This is the aforementioned thing every bit asking for a solution to the post-obit system of equations. \[\begin{array}{c} ten+y=a \\ x+2y=b \end{assortment}\nonumber \] Fix the augmented matrix and row reduce. \[\left [ \begin{array}{rr|r} ane & 1 & a \\ 1 & 2 & b \finish{array} \right ] \rightarrow \left [ \brainstorm{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \characterization{ontomatrix}\] You tin can run across from this indicate that the system has a solution. Therefore, we accept shown that for whatsoever \(a, b\), there is a \(\left [ \begin{array}{c} 10 \\ y \end{assortment} \right ]\) such that \(T\left [ \brainstorm{assortment}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \correct ]\). Thus \(T\) is onto.
At present we want to know if \(T\) is one to one. By Suggestion \(\PageIndex{1}\) it is enough to bear witness that \(A\vec{x}=0\) implies \(\vec{10}=0\). Consider the system \(A\vec{x}=0\) given by: \[\left [ \brainstorm{array}{cc} i & one \\ i & 2\\ \end{array} \correct ] \left [ \begin{array}{c} x\\ y \cease{array} \right ] = \left [ \begin{assortment}{c} 0 \\ 0 \end{assortment} \right ]\nonumber \]
This is the same as the organization given by
\[\brainstorm{array}{c} 10 + y = 0 \\ x + 2y = 0 \end{array}\nonumber \]
We need to bear witness that the solution to this system is \(x = 0\) and \(y = 0\). By setting up the augmented matrix and row reducing, nosotros stop upwardly with \[\left [ \begin{array}{rr|r} i & 0 & 0 \\ 0 & 1 & 0 \end{array} \right ]\nonumber \]
This tells united states of america that \(10 = 0\) and \(y = 0\). Returning to the original organization, this says that if
\[\left [ \begin{array}{cc} i & i \\ 1 & ii\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \finish{array} \right ]\nonumber \]
and so \[\left [ \begin{array}{c} x \\ y \end{assortment} \correct ] = \left [ \brainstorm{assortment}{c} 0 \\ 0 \stop{array} \correct ]\nonumber \]
In other words, \(A\vec{x}=0\) implies that \(\vec{x}=0\). By Suggestion \(\PageIndex{one}\), \(A\) is one to i, and and then \(T\) is also one to one.
We also could have seen that \(T\) is i 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 \(ten=2a-b\) and \(y=b-a\). Therefore, in that location is merely i vector, specifically \(\left [ \begin{assortment}{c} x \\ y \cease{array} \correct ] = \left [ \begin{array}{c} 2a-b\\ b-a \end{array} \correct ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \correct ] =\left [ \begin{array}{c} a \\ b \end{array} \correct ]\). 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 \[T \left [ \brainstorm{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \stop{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \cease{array} \right ] \in \mathbb{R}^four\nonumber \] Evidence that \(T\) is onto merely not i to one.
Solution
You can evidence that \(T\) is in fact linear.
To show that \(T\) is onto, permit \(\left [ \begin{array}{c} x \\ y \stop{array} \right ]\) be an arbitrary vector in \(\mathbb{R}^2\). Taking the vector \(\left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{assortment} \right ] \in \mathbb{R}^4\) nosotros take \[T \left [ \begin{assortment}{c} ten \\ y \\ 0 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} ten + 0 \\ y + 0 \end{array} \right ] = \left [ \begin{array}{c} x \\ y \end{array} \right ]\nonumber \] This shows that \(T\) is onto.
Past Proposition \(\PageIndex{1}\) \(T\) is ane to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). Observe that \[T \left [ \brainstorm{assortment}{r} one \\ 0 \\ 0 \\ -1 \end{array} \right ] = \left [ \brainstorm{array}{c} 1 + -1 \\ 0 + 0 \finish{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \] At that place exists a nonzero vector \(\vec{x}\) in \(\mathbb{R}^iv\) such that \(T(\vec{x}) = \vec{0}\). It follows that \(T\) is non 1 to one.
The higher up 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 data.
Allow \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation induced past the \(m \times n\) matrix \(A\). So \(T\) is i to one if and simply if the rank of \(A\) is \(n\). \(T\) is onto if and only if the rank of \(A\) is \(g\).
Consider Case \(\PageIndex{2}\). Above nosotros showed that \(T\) was onto just not 1 to one. We can at present use this theorem to make up one's mind this fact nigh \(T\).
Let \(T: \mathbb{R}^4 \mapsto \mathbb{R}^2\) exist a linear transformation divers by \[T \left [ \begin{assortment}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{assortment}{c} a + d \\ b + c \cease{array} \correct ] \mbox{ for all } \left [ \brainstorm{array}{c} a \\ b \\ c \\ d \end{assortment} \correct ] \in \mathbb{R}^iv\nonumber \] Prove that \(T\) is onto but not one to one.
Solution
Using Theorem \(\PageIndex{1}\) nosotros can show that \(T\) is onto only not ane to one from the matrix of \(T\). Recall that to discover the matrix \(A\) of \(T\), nosotros apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). The issue is the \(2 \times 4\) matrix A given by \[A = \left [ \begin{array}{rrrr} i & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{assortment} \right ]\nonumber \] Fortunately, this matrix is already in reduced row-echelon form. The rank of \(A\) is \(ii\). Therefore by the above theorem \(T\) is onto but non ane to i.
Recall that if \(Due south\) and \(T\) are linear transformations, nosotros tin can discuss their composite denoted \(S \circ T\). The following examines what happens if both \(S\) and \(T\) are onto.
Let \(T: \mathbb{R}^one thousand \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^chiliad\) exist linear transformations. If \(T\) and \(S\) are onto, and then \(S \circ T\) is onto.
Solution
Permit \(\vec{z}\in \mathbb{R}^m\). 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 \(T(\vec{10})=\vec{y}\). Thus \[\vec{z} = S(\vec{y}) = Due south(T(\vec{10})) = (ST)(\vec{ten}),\nonumber \] showing that for each \(\vec{z}\in \mathbb{R}^m\) there exists and \(\vec{x}\in \mathbb{R}^k\) such that \((ST)(\vec{10})=\vec{z}\). Therefore, \(S \circ T\) is onto.
The adjacent example shows the aforementioned concept with regards to ane-to-ane transformations.
Let \(T: \mathbb{R}^thou \mapsto \mathbb{R}^due north\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Evidence that if \(T\) and \(S\) are one to one, and so \(S \circ T\) is one-to-1.
Solution
To prove that \(S \circ T\) is one to i, we demand to show that if \(South(T (\vec{5})) = \vec{0}\) it follows that \(\vec{v} = \vec{0}\). Suppose that \(S(T (\vec{v})) = \vec{0}\). Since \(S\) is one to one, information technology follows that \(T (\vec{five}) = \vec{0}\). Similarly, since \(T\) is ane to 1, it follows that \(\vec{v} = \vec{0}\). Hence \(S \circ T\) is one to i.
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Source: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_%28Kuttler%29/05:_Linear_Transformations/5.05:_One-to-One_and_Onto_Transformations
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