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When a column vector in a matrix is a made up of "combination" of its other column vectors, it is said to be linearly dependant. Say...

$ A=\begin{bmatrix} 2 & 1 & 0\\ 4 & 5 & -6\\ 3 & 1 & 1 \end{bmatrix} $ $ 1\begin{bmatrix} 2\\ 4\\ 3 \end{bmatrix}-2\begin{bmatrix} 1\\ 5\\ 1 \end{bmatrix}=\begin{bmatrix} 0\\ -6\\ 1 \end{bmatrix} $

Otherwise, it is linearly independent. And being linearly dependent, it has the properties of being a singular matrix and therefore may have an infinite solutions or no solutions at all depending on the result matrix. Then being linearly independent, the matrix is more often a good matrix that can span the entire $R^{n}$ space and has a unique solution to its system of equations.

Then I just thinking what happens if a row vector in a matrix is made up of "combination" of its other row vectors? Say...

$P=\begin{bmatrix} 2 & 5 & 1\\ 12 & 13 & 3\\ 8 & 3 & 1 \end{bmatrix} $ $ 2\begin{bmatrix} 2 & 5 & 1 \end{bmatrix}+1\begin{bmatrix} 8 & 3 & 1 \end{bmatrix}=\begin{bmatrix} 12 & 13 & 3 \end{bmatrix} $

Does a matrix have any special properties too if its row vectors are linearly independent and linearly dependent?

Thanks!

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    ahh... I see now... Thanks InterestedGuest and mac for the help! :)2011-06-27

1 Answers 1

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Yes. The following are equivalent for a square matrix $A$:

  1. $A$ is non-singular

  2. the rows of $A$ are linearly independent

  3. the columns of $A$ are linearly independent

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    @xEnOn: leonbloy's comment would still apply: The row rank and column rank are the same.2011-06-27