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Can a non-square matrix have a full rank?

I always see cases with square matrix with full rank but seldom with non-square matrix. Can anyone help on this?

For example, is the following matrix full rank?

 A =( 1 3 10) ( 2 3 14)

My lecture slide says this does not have a full rank because any multiple of x'=[2 1 -1/2] will give Ax=0

 (1 3 10)( 2 ) ( 0 ) (2 3 14)( 1 )==( 0 ) (-1/2)

I don't think this is correct but may I check?

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1 Answer

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If a matrix is $m \times n$, then we say it has full row rank if the rank is at least $m$ and it has full column rank if the rank is at least $n$. Unless the matrix is square, it is impossible for both to occur.

We could say that the matrix is "full rank" if the rank is $\min \{ m,n \}$. I would understand this usage, even though I don't think I've actually seen it in practice.

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