Trivial solution equivalent to a solution existing

February 09, 2020

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

Suppose nn is a positive integer and Ai,jFA_{i,j} \in \mathbb{F} for i,j=1,,ni, 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 x1==xn=0x_1 = \dots = x_n = 0 is the only solution to the homogenous system of equations

k=1nA1,kxk=0k=1nAn,kxk=0.\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 c1,,cnFc_1, \dots, c_n \in \mathbb{F}, there exists a solution to the system of equations

k=1nA1,kxk=c1k=1nAn,kxk=cn.\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 AxAx, where AFn,nA \in \mathbb{F}^{n,n} and xFnx \in \mathbb{F}^n. The matrix AA induces the linear map TAL(Fn)T_A \in \mathcal{L}(\mathbb{F}^n) where TA(x)=AxT_A(x) = A x.

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

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