A First Course in Linear Algebra, 2017a

5.5. One to One and Onto Transformations 291
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.
Theorem 5.34: Matrix of a One to One or Onto Transformation
Let T : R n ↦→ R m be a linear transformation induced by the m × n matrix A. ThenT 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 5.33. Above we showed that T was onto but not one to one. We can now use this
theorem to determine this fact about T .
Example 5.35: An Onto Transformation
Let T : R 4 ↦→ R 2 be a linear transformation defined by
⎡ ⎤
⎡
a
[ ]
T ⎢ b
⎥ a + d
⎣ c ⎦ = for all ⎢
b + c ⎣
d
Prove that T is onto but not one to one.
a
b
c
d
⎤
⎥
⎦ ∈ R4
Solution. Using Theorem 5.34 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 ⃗e i of R 4 . The result is the
2 × 4matrixAgivenby
[ ]
1 0 0 1
A =
0 1 1 0
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.
♠
Recall that if S and T are linear transformations, we can discuss their composite denoted S ◦ T. The
following examines what happens if both S and T are onto.
Example 5.36: Composite of Onto Transformations
Let T : R k ↦→ R n and S : R n ↦→ R m be linear transformations. If T and S are onto, then S ◦ T is onto.
Solution. Let⃗z ∈ R m .SinceS is onto, there exists a vector⃗y ∈ R n such that S(⃗y)=⃗z. Furthermore, since
T is onto, there exists a vector⃗x ∈ R k such that T (⃗x)=⃗y. Thus
⃗z = S(⃗y)=S(T(⃗x)) = (ST)(⃗x),
showing that for each⃗z ∈ R m there exists and⃗x ∈ R k such that (ST)(⃗x)=⃗z. Therefore, S ◦ T is onto.
♠
The next example shows the same concept with regards to one-to-one transformations.