I am having trouble with the dozens and dozens of rules for determining dependence, independence and generating sets and consistent and inconsistent. I know that these are all very closely related though and I know that generating sets and independent are also very close.

I have a list of rules for independence that I wrote during class but I am not sure if they are correct.

For matrix A nxk

A) It is independent if rank of A is equal to k

B) If n > k it can't span $R^n$, I am not sure what this means.

C) If k > n then it is linearly dependent (not "and non zero solution)

So from this I think I can determine that a matrix is independent if in the reduced row echelon form of a square matrix each column has a pivot. Is this wrong?

For example I have a square matrix (rows = columns)

\begin{bmatrix} 1 & -1 & 1 \\[0.3em] -1 & 0 & 2 \\[0.3em] -2 & 1& 1 \end{bmatrix}

To get reduced row echelon form I do the following transformations:

$R_1 +R_2$

\begin{bmatrix} 1 & -1 & 1 \\[0.3em] 0 & -1 & 3 \\[0.3em] -2 & 1& 1 \end{bmatrix} $2R_1+R_3$

\begin{bmatrix} 1 & -1 & 1 \\[0.3em] 0 & -1 & 3 \\[0.3em] 0 & 0 & 2 \end{bmatrix}

And from here it is fairly trivial to get it into reduced row echelon, so my rank is 3 and my columns are 3. Why is it not independent?


You made a small mistake in the last step: adding twice the first row to the third should give you $$\left[\begin{array}{ccc}1 & -1 & 1\\0 & -1 & 3\\ 0 & -1 & 3\end{array}\right]$$ and you should begin to see how the row-reduction will end.

  • $\begingroup$ Ok, but more important are my assumption correct? $\endgroup$ – Paul the Pirate Oct 2 '13 at 19:45
  • $\begingroup$ Yes. All of the pivots are nonzero if and only if the matrix has rank $\min(n,k)$. $\endgroup$ – user7530 Oct 2 '13 at 19:47
  • $\begingroup$ But I mean my A, B and C method for finding dependence or not. $\endgroup$ – Paul the Pirate Oct 2 '13 at 19:58
  • $\begingroup$ I'm not sure. "linearly dependent" is not too precise. An $n\times k$ matrix has linearly independent rows if its rank is $n$ and linearly independent columns if its rank is $k$. The columns span a linear subspace of dimension equal to the rank. This space is all of $\mathbb{R}^n$ when the rank is equal to $n$. $\endgroup$ – user7530 Oct 2 '13 at 20:02

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