If a vector in the set of vectors $$\{v_1,v_2,v_3...v_n\}$$ can be written as a linear combination of other vectors, must a row in the matrix formed by linearly combining the vectors be equal to a linear combination of other rows? Or, in other words, does column linear dependence imply row linear dependence and vice versa?

Since people talk of linear systems being 'linearly dependent', I would think that the answer to my questions would be yes. However, this example makes me think otherwise:

$$\left(\begin{array}{c} 4 & 1 & 5 \\ -2 & 7 & 6 \\ 2 & 8 & 11 \end{array}\right)$$

The third row is equal to the sum of the first two rows, so the rows are linearly dependent. The first vector is not, however, a linear combination of the other vectors. The second vector is not a linear combination of the others. The third vector is not a linear combination of the others. No vector is a scalar multiple of another. Unless those statements are false, the columns must be linearly independent despite the rows being linearly dependent.

  • 4
    $\begingroup$ This question appears to be off-topic because it is about math. $\endgroup$
    – bmargulies
    Oct 1, 2013 at 23:26

1 Answer 1


The column rank of a matrix is the number of linearly independent columns. The row rank can be defined the same way. Row rank and column rank are the same, wich is why that number is usually just called the rank.

The rows of a matrix are linearly independent if the number of rows equals the rank. The same holds for the columns. This means that for a square matrix, row independence and column independence are the same, and likewise dependence. For a non-square matrix, you will always have dependence along one direction. If the rank still agrees with the smaller of the two dimensions of the matrix, then you have indepedendence in that direction, and the matrix is said to have full rank.

As for your matrix, you said that for row vectors you have

$$w_3 = w_1+w_2$$

For the column vectors, you have


Not so easy to see at first glance, but easy to compute from a system of two linear equations, e.g.

$$\left. \begin{array}{r@{}l@{}r@{}c@{}r} \lambda & + & 5\mu & = & 4 \\ 7\lambda & + & 6\mu &= & -2 \end{array}\quad\right\} \quad\Longrightarrow\quad \left\{ \begin{array}{r@{}c@{}r} \lambda &=& -\tfrac{34}{29} \\ \mu &=& \tfrac{30}{29} \end{array} \right. $$

You can use Wolfram Alpha to solve this is you want to.

  • $\begingroup$ +1. Too fast for me today. I was just writing this answer, although I was going to write v3 as a linear combination of v1 and v2. $\endgroup$
    – user1608
    Oct 1, 2013 at 23:40

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