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# Is this function injective?

## Problem

Let

$$
A = \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 2 & 2 \\
2 & 2 & 2
\end{bmatrix}\text{ and }
B = \begin{bmatrix}
1 & 1 & 2 \\
1 & 2 & 2
\end{bmatrix}.
$$

Define funcitons $f_A: \mathbb{R}^3\rightarrow\mathbb{R}^4$ by $\bx\mapsto A\bx$ and $f_B: \mathbb{R}^3\rightarrow\mathbb{R}^2$ by $\bx\mapsto B\bx$. For each of $f_A$ and $f_B$, determine if it is injective.

## Thought

As usual, the definition is important: A function $f$ is injective if for any $x_1\neq x_2$ in the domain, it must be $f(x_1)\neq f(x_2)$. Alternatively, you could say whenever $f(x_1) = f(x_2)$ for some $x_1,x_2$ in the domain, we have $x_1 = x_2$.

## Sample answer

First we consider $f_A$. Let $\bx_1,\bx_2\in\mathbb{R}^3$. Suppose $f_A(\bx_1) = f(\bx_2)$. By definition, this means $A\bx_1 = A\bx_2$ and $A(\bx_1 - \bx_2) = \bzero$. By doing the row operations, $A$ has three leading variables and no free variable, so $\ker(A) = \{\bzero\}$. Therefore, $A(\bx_1 - \bx_2) = \bzero$ implies $\bx_1 - \bx_2 = \bzero$ and $\bx_1 = \bx_2$. Thus, $f_A$ is injective.

For $f_B$, it is not hard to see that when

$$
\bx_1 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \text{ and }
\bx_2 = \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix},
$$

we have $f_B(\bx_1) = f_B(\bx_2) = \bzero$, so $f_B$ is not injective.

## Note

Let $A$ be an $m\times n$ matrix and define $f_A: \mathbb{R}^n\rightarrow\mathbb{R}^m$ by $\bx\mapsto A\bx$. Then you may show that the following are equivalent:

1. $f_A$ is injective.
2. $\ker(A) = \{\bzero\}$.
3. $\nul(A) = 0$.
4. $\rank(A) = n$.

Think about why they mean the same thing.

*This note can be found at Course website > Learning resources.*