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Linear Algebra: What makes the column space and row space of a matrix have the same dimension?

  • Why do the column space and row space of a matrix have the same dimension I know the rank is shared by the column and row space, but how can I understand this intuitively?

  • Answer:

    First, a light-weight proof, in case that's intuitive enough: Let's say matrix A is m x n. A has n columns, each of which are m-dimensional vectors. Let's say the column space of A is c-dimensional. c may be less than m and n. There is a basis of c vectors (each m-dimensional) that spans the column space of A. So the columns of A can be written in terms of these c vectors. To express that, write the matrix B, containing those c vectors as columns. Then we'll have A = BC, where C's columns are the coordinates of columns of A in terms of this basis. This is the key point -- won't explain it here at length but it's important in what's next. (We don't care what C is for purposes here; it exists. Same for B.) Now turn back but along a different path. We could also view A = BC as a statement about the basis for A's rows. B's rows are coordinates for A's rows expressed in the basis of C's rows. C has c rows. A's row space is spanned by these c vectors. That doesn't quite mean the space is c-dimensional. It could be less; the rows of C may not be linearly independent. But at least we know that dim(row space of A) <= dim(col space of A) Now apply to same logic to the transpose of A. dim(row space of A') <= dim(col space of A') ,  so dim(col space of A) <= dim(row space of A) Since the two dimensions are <= each other, they must be equal. Now the most intuitive observation I know to make: B has c columns and C has c rows of course. This is the nature of the symmetry that leads in two directions to conclude the same thing about the row and column rank. B is coordinates in terms of C, C is coordinates in terms of B. This shared c-dimensional space gives rise to the row/column space both.

Sean Owen at Quora Visit the source

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Show that both the row rank and the column rank of an [math]m \times n[/math] matrix [math]A[/math] can be characterised as the smallest positive integer [math]r[/math] for which there exists an [math]m \times r[/math] matrix [math]B[/math] and an [math]r \times n[/math] matrix [math]C[/math] such that [math]A = BC[/math]. After giving it some thought, this one is more intuitive to me. Multiplying the matrix on the left by an invertible matrix doesn't change its column rank because this operation corresponds to applying an invertible linear transformation to the image of the linear transformation defined by the matrix. Thus elementary row reduction doesn't change the column rank of the matrix, and neither does permuting the rows. It's clear (at least to me) that these operations do not change the row rank either. One can similarly argue that the row rank of the matrix is unaffected by elementary column reductions and column permutations. Now change the matrix to row-reduced echelon form as well as column-reduced echelon form, and apply some permutations too, to finally get a matrix which is of the form [math]\begin{pmatrix}I_r & 0_{r \times (n-r)}\\ 0_{(n-r)\times r} &0_{(n-r)\times(n-r)}\end{pmatrix}[/math]. This [math]r[/math] is clearly equal to both the row rank and the column rank of the new matrix, and hence of the old matrix.

Anurag Bishnoi

I think has given a great answer. I'm just trying to give a more intuitive, IHMO, and informal one. The following reasoning is based on the statement that "In a D dimensional space, we need D scalars to present a vector." Let's say matrix A is m * n. A's column rank is R_c. This means that you need R_c scalars to represent a vector in column space.  Therefore, for the m elements of each column vector, we need R_c of them.  In other words, for each column vector, we need to choose R_c row indices. But we don't know if we can find R_c row indices that work for all the column vectors: if we can then there are R_c independent row vectors, if we cannot there are more than R_c independent row vectors. Therefore, R_r >= R_c. Considering A', we know R_c >= R_r. Therefore, R_c == R_r.

Gavin Weiguang Ding

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