The row space and column space of an mbyn matrix with real entries is the subspace of R^{n} generated by the row vectors and column vectors, respectiviely, of the matrix. Its dimension is equal to the rank of the matrix and is at most min(m,n).^{[1]}
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Overview
Let A be a n by m matrix. Then
If one considers the matrix as a linear transformation from R^{n} to R^{m}, then the column space of the matrix equals the image of this linear transformation.
The column space of a matrix A is the set of all linear combinations of the columns in A. If A = [a_{1}, ...., a_{n}], then colsp(A) = span {a_{1}, ...., a_{n}}.
The concept of row space generalises to matrices to C, the field of complex numbers, or to any field.
Intuitively, given a matrix A, the action of the matrix A on a vector x will return a linear combination of the columns of A weighted by the coordinates of x as coefficients. Another way to look at this is that it will (1) first project x into the row space of A, (2) perform an invertible transformation, and (3) place the resulting vector y in the column space of A. Thus the result y =A x must reside in the column space of A. See the singular value decomposition for more details on this second interpretation.
Example
Given a matrix J:
the rows are r_{1} = (2,4,1,3,2), r_{2} = (−1,−2,1,0,5), r_{3} = (1,6,2,2,2), r_{4} = (3,6,2,5,1). Consequently the row space of J is the subspace of R^{5} spanned by { r_{1}, r_{2}, r_{3}, r_{4} }. Since these four row vectors are linearly independent, the row space is 4dimensional. Moreover in this case it can be seen that they are all orthogonal to the vector n = (6,−1,4,−4,0), so it can be deduced that the row space consists of all vectors in R^{5} that are orthogonal to n.
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