Linear maps are essential in linear algebra, and 1-to-1 and onto linear maps are especially important. In this post, we investigate 1-to-1 and onto linear maps and discuss them with suitable examples.

#### 1-to-1 linear maps

A function is, by definition, 1-to-1 (also called one-to-one or injective) if \(f(x) = f(y)\) implies \(x=y\), for all \(x\) and \(y\) in the defined domain of the function \(f\). It may happen that a function on \(A\) is not 1-to-1 but on a smaller set \(B\) is 1-to-1.

In linear algebra there is an excellent criterion to check if a linear map (also called a linear function or linear transformation) is 1-to-1 or not. For example, \(f(x) = x^2\) is not 1-to-1 on \(\mathbb{R}\) but it is 1-to-1 on \(\mathbb{R}^{+}\) (the set of positive real numbers).

**Result**. A linear map \(f\) from a vector space \(V\) to a vector space \(W\) is 1-to-1 if and only if the kernel of the linear map \(f\) is zero, i.e., \(\ker(f) = \{0\}\).

The proof of the above result is as follows:

Let \(f\) be a 1-to-1 (injective) linear map. Assume that \(u\) is a vector in the kernel of \(f\), then \(f(u) = 0\), by definition. Since \(f\) is linear, \(f(0) = 0\). This implies that \(f(u) = f(0)\). Now, since \(f\) is 1-to-1, we obtain that \(u\) is the zero vector and this means that the kernel of \(f\) vanishes.

Conversely, assume that the kernel of the linear map \(f\) is zero. In order to prove that \(f\) is 1-to-1, we assume that \(f(u) = f(v)\). So, \(f(u) – f(v) = 0\). Since \(f\) is linear, we have \(f(u – v) = 0\). Now, since \(\ker(f) = \{0\}\), we deduce that \(u – v = 0\) and this implies that \(u = v\). Hence, \(f\) is 1-to-1.

**Exercise**. Prove that a linear map $$f: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ is 1-to-1 if and only if the columns of its matrix are linearly independent.

#### Onto linear maps

Let us recall that a function \(f\) from \(A\) to \(B\) is, by definition, onto if for each \(b\in B\) there is an \(a \in A\) such that \(f(a) = b\), i.e. each element in the codomain is mapped from an element in the domain. It is clear that the function is onto if and only if \(f(A) = B\).

Now, let \(f\) be a linear map from \(\mathbb{R}^n\) to \(\mathbb{R}^m\). The map \(f\) is onto if and only if \(f(\mathbb{R}^n) = \mathbb{R}^m\). By the rank-nullity theorem in linear algebra discussed in column space and rank, we have $$\hbox{rank}(A) + \hbox{nullity}(A) = n.$$ It is evident that \(f\) is onto if and only if the dimension of the column space of \(f\) is \(m\). Hence, we have proved the following criterion for onto linear maps over finite dimensional real vector spaces.

**A criterion for onto linear maps**. A linear map \(f\) from \(\mathbb{R}^n\) to \(\mathbb{R}^m\) is onto if and only if the dimension of \(\ker(f)\) is \(n-m\).

### Examples of 1-to-1 and onto linear maps

Let \(f: \mathbb{R}^2 \rightarrow \mathbb{R}^3\) be a function defined by $$f(x,y) = (x+y, x-y, 1).$$ This function is 1-to-1 (prove this). However, it is not a linear map because $$f(0,0) \neq (0,0,0).$$

Let \(g: \mathbb{R}^2 \rightarrow \mathbb{R}^3\) be a function defined by $$g(x,y) = (x+y, x-y, 0).$$ This is a 1-to-1 linear map.

Let \(h: \mathbb{R}^3 \rightarrow \mathbb{R}^2\) be a function defined by $$g(x,y,z) = (x+y, x-y).$$ This function is a linear map but it cannot be 1-to-1, because by the rank-nullity theorem, we have the following result and the reader can prove this as an exercise:

**Exercise**. If a linear map \(f: \mathbb{R}^n \rightarrow \mathbb{R}^m\) is 1-to-1, then \(n \leq m\).

Let \(f: \mathbb{R}^4 \rightarrow \mathbb{R}^2\) be a function defined by $$f(x,y,z,t) = (x,0).$$ Then, \(f\) is linear but not onto (and of course, not 1-to-1).

Let \(g: \mathbb{R}^4 \rightarrow \mathbb{R}^2\) be a function defined by $$g(x,y,z,t) = $$ $$(x+z,x-z).$$ Then, \(g\) is an onto linear map but of course not a 1-to-1 linear map.

**Exercise**. Let \(f: \mathbb{R}^n \rightarrow \mathbb{R}^n\) be a linear map. Prove that the following statements are equivalent:

- \(f\) is 1-to-1 (injective).
- \(f\) is onto (surjective).
- The matrix of \(f\) is invertible.