A wide matrix is full rank but its columns are not linearly dependent as expected. Why?

Question

I have a 2×3 matrix,

\begin{pmatrix}1 & 2 & 4 \\ 2 & 3 & 5 \end{pmatrix}

and its rank from row echelon form is 2 which is also the maximum possible rank for the matrix, i.e., a full rank matrix. This means the matrix has two linearly independent rows and so the set of vectors that the matrix represents is linearly independent. But a wide matrix should have linearly dependent columns. See, MathsStackExchange

So while one concept is hinting at linear dependency the other concept is hinting at linear independency. Is this a contradiction? Also the three vectors are non-collinear and the span of any two vectors covers the third vector too. Therefore the addition of the third vector does not increases the span, and hence from the increasing span criterion, the set of vectors should be linearly dependent. But I am finding the matrix to be of full rank as explained above.

Answer 1

If we label the columns $c_1,c_2$ and $c_3$ note that $-2c_1+3c_2-c_3=0$ so they are linearly dependent. However for the rows $r_1$ and $r_2$ there is no scalar $\alpha$ such that $r_1+\alpha r_2=0$ so they are linearly independent. In both cases two of the vectors are sufficient to span the column and row space. In general the row and column space will have dimensions equal to the rank of the matrix.

Answer 2

The rank corresponds to the dimension of $C(A)$ and $C(A^T)$ that is to the maximum number of independent columns and independent rows. In this case the matrix has $2$ independent rows but columns are only independent in pairs (any pair spans $\mathbb R^2$).

Note that for this other case

\begin{pmatrix}1 & 2 & 4 \\ 2 & 4 & 5 \end{pmatrix}

rank is also $2$ but firs and second colums are multiple one of each other as we can argue from the zero column we obtain by the ERRF.