Question

Suppose A is an n×n matrix, and let v1,.....vn belong to R^n. Suppose {Av1,.....Avn} is linearly independent prove that A is non singular

Answer 1

$\begin{aligned} &\text { Given that }\left\{A v_{1}, A v_{2}, A v_{3} \ldots A v_{n}\right\} \text { is linearly } \\ &\text { independent vectors of } \mathbb{R}^{n} \\ &\text { As } \mathbb{R}^{n} \text { has dimension } n \text {, this means that } \\ &\left\{A v_{1}, A v_{2}, A v_{3} \ldots A v_{n}\right\} \text { must be a basis for } \mathbb{R}^{n} \\ &\text { Therefore, the column space of } A \text {, which consists of } \\ &\text { vectors of the form }\left\{A v: v \in \mathbb{R}^{n}\right\} \text { is of dimension } n \text {. } \\ &\text { In other words, rank }(A)=n \\ &\text { We want to show that } A \text { is non-singular } \\ &\text { That is, } x \in \mathbb{R}^{n} \text { and if } A x=0 \Rightarrow x=0 \\ &\text { Let } A x=0 \text { for some } x \in \mathbb{R}^{n} \\ &\text { Thus, we must have } x \in \mathrm{ker}(A) \text { (the null space of the } \\ &\text { given matrix) which is defined as } \\ &\text { ker }(A)=\left\{x \in \mathbb{R}^{n}: A x=0\right\} \text { and } \\ &\text { null }(A)=\operatorname{dim}(\text { ker }(A)) \\ &\text { As } A: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} \text { by Rank-Nullity theorem, we must } \\ &\text { have } \\ &\text { rank } A+\text { null } A=n \\ &\text { But rank } A=n \text { so that } \\ &n+\text { null } A=n \Rightarrow \text { null } A=0 \\ &\text { Thus, we must have } \\ &\text { ker }(A)=\left\{x \in \mathbb{R}^{n}: A x=0\right\}=\{0\} \text { as this is the } \\ &\text { only subspace having rank } 0 \\ &\text { Therefore, } \\ &x \in \text { ker }(A)=\left\{x \in \mathbb{R}^{n}: A x=0\right\}=\{0\} \Rightarrow x=0 \\ &\text { We have shown that for all } x \in \mathbb{R}^{n}, \text { we have } \\ &A x=0 \Rightarrow x=0 \\ &\text { Therefore, the matrix } A \text { is non-singular. } \end{aligned}$