# confusing between injective and surjective

Given a linear operator $\mathcal{A} \colon \mathbb{R}^{n} \to \mathbb{R}^{m}$.

1. Please check if I understand correct:

• $\mathcal{A}$ is injective if and only if its associated matrix has full column rank, which equivalent further to the fact that $\mathcal{A}^{*} \mathcal{A}$ is positive definite: $$\label{1}\tag{1} \left\langle \mathcal{A}^{*} \mathcal{A} x , x \right\rangle \geq \lambda_{\min} \left( \mathcal{A}^{*} \mathcal{A} \right) \left\lVert x \right\rVert ^{2} > 0 , \forall x \in \mathbb{R}^{n} .$$
• $\mathcal{A}$ is surjective if and only if its associated matrix has full row rank, which equivalent further to the fact that $\mathcal{A} \mathcal{A}^{*}$ is positive definite: $$\label{2}\tag{2} \left\langle \mathcal{A} \mathcal{A}^{*} y , y \right\rangle \geq \lambda_{\min} \left( \mathcal{A} \mathcal{A}^{*} \right) \left\lVert y \right\rVert ^{2} > 0 , \forall x \in \mathbb{R}^{m} .$$ here $\lambda_{\min}$ denotes the smallest eigenvalue of the associated matrix.
2. If everything is correct, does it means that an example of an injective but not surjective operator is a matrix which has full column rank but not full row rank and vice versa?

3. Is there any easy way to avoid this confusion? As I always mesh up between column and row. Any ideas are appreciated.