7Conclusion

We have seen that there is a 1-1 correspondence, induced by the matrix, between the subspace of $\R^n$ spanned by its rows and the subspace of $\R^m$ spanned by its columns (that is, the image of the matrix). This means that the dimensions of these two subspaces are equal. The dimension of the space spanned by the rows is the row rank and the dimension of the space spanned by the columns is the column rank. We have seen before why they are equal. Because of that one simply speaks of the “rank” of the matrix.

Our derivation of the equality of row rank and column rank works only in $\R^n$, since our argument relies on the standard scalar product in $\R^n$: $\langle x,y\rangle =x^Ty$. In $\mathbb{C}^n$ the mapping $\langle x,y\rangle =x^Ty$ is not a scalar product, since in $\mathbb{C}^2$

Note that this means that the vector $(1,i)^T$ is in the kernel of the matrix

Hence, even over $\mathbb C$, the kernel of a matrix is in general not the orthogonal complement of the rows.

If $x^Ty$ was a sensible notion of a scalar product on $\mathbb{C}^2$, this would mean that the nonzero vector $(1, i)^T$ is orthogonal to itself. Intuitively this does not make sense.