## Trivial solution equivalent to a solution existing

#### February 09, 2020

From Axler's Linear Algebra Done Right, chapter 3.D, exercise 20:

Suppose $n$ is a positive integer and $A_{i,j} \in \mathbb{F}$ for $i, j = 1, \dots, n$. Prove that the following are equivalent. (note that in both parts below, the number of equations equals the number of variables):

(a) The trivial solution $x_1 = \dots = x_n = 0$ is the only solution to the homogenous system of equations

\begin{aligned} \sum^n_{k=1} A_{1,k}x_k&=0\\ \vdots\\ \sum^n_{k=1} A_{n,k}x_k&=0. \end{aligned}

(b) For every $c_1, \dots, c_n \in \mathbb{F}$, there exists a solution to the system of equations

\begin{aligned} \sum^n_{k=1} A_{1,k}x_k&=c_1\\ \vdots\\ \sum^n_{k=1} A_{n,k}x_k&=c_n. \end{aligned}

The left hand side of both system of equations can be rewritten as the matrix equation $Ax$, where $A \in \mathbb{F}^{n,n}$ and $x \in \mathbb{F}^n$. The matrix $A$ induces the linear map $T_A \in \mathcal{L}(\mathbb{F}^n)$ where $T_A(x) = A x$.

Condition (a) is equivalent to $\text{null } T_A = \{0\}$, which is equivalent to $T_A$ is injective. Condition (b) is equivalent to $\text{range } T_A = \mathbb{F^n}$, which is equivalent to $T_A$ is surjective.

Hence, [Axler 3.69] implies that injectivity and surjectivity are equivalent for linear operators, and as $T_A \in \mathcal{L}(\mathbb{F}^n)$, we have $T_A$ injective iff $T_A$ surjective. Therefore, (a) and (b) are equivalent.