3.2 One-to-one and Onto Transformations¶
Domain and Codomain¶
A transformation
based on the \(n\times m\) matrix \(A\) always maps from the domain \(\mathbb R^m\) into the range \(\mathbb R^n\). The vectors \(\vec x\) in the domain, to be able to be multiplied in the order \(A\vec x\) must have the same number of components as the columns of \(A\), which is \(m\). The vectors \(\vec y\) in the range will have \(n\) components because \(n\) dot products will be needed.
We reproduce a theorem from Margalit to show what we can do with MATLAB to verify a matrix transformation is one-to-one.
Theorem. One-to-one matrix transformations.¶
Let \(A\) be an \(m\times n\) matrix, and let \(T\left(\vec x\right) = A \vec x\) be the associated matrix transformation. The following statements are equivalent:
\(T\) is one-to-one.
For every \(\vec b \in \mathbb R^m\), the equation \(T\left(\vec x\right) = \vec b\) has at most one solution.
For every \(\vec b\in \mathbb R^n\), the equation \(A\vec x = \vec b\) has a unique solution or is inconsistent.
\(A\vec x = \vec 0\) has only the trivial solution.
The columns of \(A\) are linearly independent.
\(A\) has a pivot in every column.
The range of \(T\) has dimension \(n\).
To verify a matrix transformation is one-to-one, we can use option 6: row-reduce \(A\) to determine if every column corresponds to a pivot position.
Examples¶
Consider the transformations corresponding to the following matrices:
It should be clear right away that the wide matrix \(D\) could never be associated with a one-to-one transformation. There is no way it could have pivots in all five columns since only three pivots are possible. Let’s check on the others.
A = [3 -11 ; 0 1 ; 0 1 ];
B = [0 -1 1 ; 2 3 5 ; -4 9 -24 ];
C = [0 0 -1 1 ; 1 4 1 1 ; 1 4 -2 6 ; 1 4 -2 4 ];
D = [4 2 23 10 8 ; -4 -2 -18 -7 -6 ; -16 -8 -57 -18 -19 ];
rref(A)
rref(B)
ans =
1 0
0 1
0 0
ans =
1 0 0
0 1 0
0 0 1
Both matrices \(A\) and \(B\) have a pivot in each column, so their corresponding transformations will be one-to-one.
rref(C)
ans =
1 4 0 0
0 0 1 0
0 0 0 1
0 0 0 0
While it certainly would be possible for a \(4\times 4\) matrix to be one-to-one, this one is not. It has only three pivots.
Theorem. Onto matrix transformations.¶
Let \(A\) be an \(m\times n\) matrix, and let \(T\left(\vec x\right) = A \vec x\) be the associated matrix transformation. The following statements are equivalent:
\(T\) is onto.
\(T\left(\vec x\right) = \vec b\) has at least one solution for every \(\vec b \in \mathbb R^m\).
\(A\vec x = \vec b\) is consistent for every \(\vec b \in \mathbb R^m\)
The columns of \(A\) span \(\mathbb R^m\).
\(A\) has a pivot in every row.
The range of \(T\) has dimension \(m\).
Let’s consider the same 4 matrices as before.
A = [3 -11 ; 0 1 ; 0 1 ];
B = [0 -1 1 ; 2 3 5 ; -4 9 -24 ];
C = [0 0 -1 1 ; 1 4 1 1 ; 1 4 -2 6 ; 1 4 -2 4 ];
D = [4 2 23 10 8 ; -4 -2 -18 -7 -6 ; -16 -8 -57 -18 -19 ];
rref(A)
ans =
1 0
0 1
0 0
Of course, there is no way for the tall matrix \(A\) to be onto since it can have at most 2 pivots.
rref(B)
ans =
1 0 0
0 1 0
0 0 1
rref(D)
ans =
1.0000 0.5000 0 0 -1.2500
0 0 1.0000 0 1.0000
0 0 0 1.0000 -1.0000
The transformations associated with matrices \(B\) and \(D\) are onto because the matrices have pivots in every row.
rref(C)
ans =
1 4 0 0
0 0 1 0
0 0 0 1
0 0 0 0
The transformation associated with matrix \(C\) is not onto because not every row has a pivot.