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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: \begin{equation} \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} . \end{equation}
    • $\mathcal{A}$ 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: \begin{equation} \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} . \end{equation} 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.

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Take A matrix of order 2*3 choose ist row arbitrarily and choose second row such that not multiple of ist row. Then row rank of this matrix is 2 and column rank is also 2. As matrix A is of some linear transformation R^3to R^2. This linear transformation cannot be one one. So for onto we have to just check given matrix is full rank or not. Again make a matrix B of order 3*2 such second column is not multiple of other. Here column rank is 2. As this matrix is of some linear transformation from R^2 to R^3. So this transformation cannot be onto.so for injective we have to check B is full column rank.

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