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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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    $\begingroup$ Dependence between the columns implies dependence between the rows, and vice-versa. Take a close look: the third row is, in fact, $1/17$ of the first row $+$ $3/17$ of the second row. $\endgroup$ Jun 27, 2011 at 18:08
  • $\begingroup$ So I can safely say that the properties that apply to a matrix which is linearly independent for its column vectors is the same as another matrix that is linearly independent for its row vectors? $\endgroup$
    – xenon
    Jun 27, 2011 at 18:12
  • $\begingroup$ Essentially yes, since those qualities are equivalent -- you can't have one without another. =) $\endgroup$ Jun 27, 2011 at 18:13
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    $\begingroup$ Any property that applies to every square matrix with linearly independent columns also applies to every square matrix with linearly independent rows. (This is a tautology once you realise that these two classes of matrices are equal). $\endgroup$
    – mac
    Jun 27, 2011 at 18:16
  • $\begingroup$ ahh... I see now... Thanks InterestedGuest and mac for the help! :) $\endgroup$
    – xenon
    Jun 27, 2011 at 18:20

1 Answer 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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    $\begingroup$ More in general: any matrix has the same row and column "rank" (amount of linearly independent rows/columns). This is a fundamental property, and it's not obvious. $\endgroup$
    – leonbloy
    Jun 27, 2011 at 18:09
  • $\begingroup$ If the matrix is not a square matrix, what properties would differ? $\endgroup$
    – xenon
    Jun 27, 2011 at 18:13
  • $\begingroup$ @xEnOn: leonbloy's comment would still apply: The row rank and column rank are the same. $\endgroup$
    – joriki
    Jun 27, 2011 at 18:17

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