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    The $m \times n$ matrices have dimension $mn$

    E-HK-23-12

    Area
    Algebra
    Topics
    Dimension, Linear Algebra
    Status
    draft

    Prove that the space of all \(m \times n\) matrices over the field \(F\) has dimension \(mn\), by exhibiting a basis for this space.

    Solution

    1. Let \(E_{ij}\) be matrix with \(1\) in \((i,j)\) and \(0\) elsewhere.

    definition.

    2. \(\{E_{ij}:1\le i\le m,1\le j\le n\}\) spans.

    any matrix \(A=(a_{ij})=\sum a_{ij}E_{ij}\).

    3. They are linearly independent: \(\sum c_{ij}E_{ij}=0\) implies \(c_{ij}=0\).

    entrywise.

    4. Hence basis, dimension \(mn\).

    2 and 3.

    5. Q.E.D.

    4.