Rank of a matrix 

Alternative definitions

The maximal number of linearly independent columns of the m-by-n matrix A with entries in the field F is equal to the dimension of the column space of A (the column space being the subspace of F^m generated by the columns of A).

Alternatively and equivalently, we can define the rank of A as the maximal number of linearly independent rows of A, which equals the dimension of the row space of A.

If one considers the matrix A as a linear map

f : Fⁿ -> F^m

f(x) = Ax

The Rank Theorem can be expressed as: rank A + dim(Nul A) = n, given an m by n matrix A.

Properties

We assume that A is an m-by-n matrix over the field F and describes a linear map f as above.

• rk(A) ≤ min(m,n)
• f is injective if and only if rk(A) = n.
• f is surjective if and only if rk(A) = m.
• In the case of a square matrix A (m = n), then A is invertible if and only if rk(A) = n.
• If B is any n-by-k matrix, then rk(AB) ≤ min(rk(A), rk(B).
• If B is an n-by-k matrix with rk(B) = n, then rk(AB) = rk(A)
• If C is an l-by-m matrix with rk(C) = m, then rk(CA) = rk(A)
• The rank of A is equal to r if and only if there exists an invertible m-by-m matrix X and an invertible n-by-n matrix Y such that

<math>

  XAY =
  \begin{bmatrix}
    I_r & 0 \\
    0 & 0 \\
  \end{bmatrix}

where I_r denotes the r-by-r identity matrix.

• The rank of a matrix plus the nullity of the matrix equals the number of columns of the matrix.

Computation

The easiest way to compute the rank of a matrix A is given by the Gauss elimination method. The row-echelon form of A produced by the Gauss algorithm has the same rank as A, and its rank can be read off as the number of non-zero rows.

Consider for example the 4-by-4 matrix

<math>

  A =
  \begin{bmatrix}
    2 & 4 & 1 & 3 \\
    -1 & -2 & 1 & 0 \\
    0 & 0 & 2 & 2 \\
    3 & 6 & 2 & 5 \\
  \end{bmatrix}

We see that the second column is twice the first column, and that the fourth column equals the sum of the first and the third. The first and the third columns are linearly independent, so the rank of A is two. This can be confirmed with the Gauss algorithm. It produces the following row echelon form of A:

<math>

  A =
  \begin{bmatrix}
    1 & 2 & 0 & 1 \\
    0 & 0 & 1 & 1 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
  \end{bmatrix}

which has two non-zero rows.

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