Rank of a Matrix

Definition 1: The rank of a matrix A, denoted rank(A), is the maximum number of independent rows in A.

Observation: Here we view each row in matrix A as a row vector. Thus rank(A) = the dimension of the span of the set of rows in A (see Definition 2 of Linear Independent Vectors). For an m × n matrix A, clearly rank(A) ≤ m.

It turns out that the rank of a matrix A is also equal to the column rank, i.e. the maximum number of independent columns in A (per Property 1). Thus if A is an m × n matrix, then rank(A) ≤ min(m, n). This also means that rank(A) = the dimension of the span of the rows in A = the dimension of the span of columns in A (see Definition 3 of Linear Independent Vectors).

Property 1: For any matrix A, rank(A) = the maximum number of independent columns in A.

Proof: Let A be an m × n matrix and let p = column rank of A, i.e. the maximum number of independent column vectors in A. Now let C₁, …, C_p be any basis for the span of the column vectors in A. Thus any column A_j in A can be expressed uniquely as a linear combination of the basis elements, and so there are b_ij such that A_i = . Thus A = CB where C is the m × p matrix whose columns are C₁, …, C_p and B = [b_ij], a p × n matrix.

Since A = CB, every row in A can be expressed as a linear combination of the rows in B, namely . This means that the span of the rows in A is a subset of the span of the rows in B, and rank(A) ≤ rank(B) ≤ p = column rank of A.

Applying a similar argument to A^T, we conclude that rank(A^T ) ≤ column rank of A^T. But clearly rank(A^T ) = column rank of A and column rank of A^T = rank(A). Thus we conclude that column rank of A ≤ rank(A). Putting both conclusions together proves that rank(A) = column rank of A.

Observation: Since each of the transformations described in Property 9 of Simultaneous Linear Equations (i.e. the transformation in Gaussian elimination) doesn’t change the number of independent rows in a matrix, the rank of matrix A is equal to the number of non-zero rows that remain after Gaussian Elimination has been applied.

Property 2: A square n × n matrix is invertible if and only if its rank is n.

Proof: As observed just prior to Example 9 in Simultaneous Linear Equations, we can use Gaussian Elimination on to invert a square n × n matrix A. The procedure produces an inverse if and only if the rank of A is n.

Observation: This means that a square matrix A is invertible if and only if its rows are linearly independent. Since a square matrix is invertible if and only if its transpose is invertible, this also means that a matrix is invertible if and only if its columns are linearly independent.

Observation: Since Gaussian Elimination preserves the rank of a matrix, we can restate the conclusions made in Simultaneous Linear Equations about solutions to simultaneous linear equations in the following property.

Property 3: For an m × n matrix A and m × 1 column vector C, If rank(A|C) = rank(A) = n then the equation AX = C has a unique solution. If rank(A|C) = rank(A) < n there are an infinite number of solutions and if rank(A|C) > rank(A) there are no solutions.

In the case where C = 0, AX = 0 has a unique solution, namely the trivial solution X = 0, when rank(A) = n and an infinite number of solutions otherwise.

Property 4: The following are equivalent for an n × n matrix A:

• A is invertible
• det A ≠ 0
• X = 0 is the only solution of AX = 0
• AX = C has a unique solution for all n × 1 vectors C
• rank(A) = n

Proof: This is a combination of Property 2 and 3 and Property 4 of Determinants and Simultaneous Linear Equations.

Observation: Note that if AX = C has a unique solution for some n × 1 column vector C, then by Property 3, rank(A) = n and so by Property 2, A is invertible. Conversely, if A is invertible then for any n × 1 column vector C,  X = A^-1C is the unique solution to AX = C.

Observation: Since an n × n matrix A is invertible if and only if rank(A) = n, it follows by Corollary 4 of Linear Independent Vectors, that the rows in A form a basis for the set of all the 1 × n row vectors. Similarly the columns in A form a basis for the set of all the n × 1 column vectors.

Property 5: For a square matrix A, if there is a matrix B such that AB = I or BA = I then A is invertible and A^-1 = B.

Proof: Assume first that there is a square matrix B such that BA = I. Now if AX = 0, then X = IX = (BA)X = B(AX) = B0 = 0. This shows that AX = 0 only has the trivial solution X = 0, which by Property 4 means that A is invertible and A^-1 exists. Thus, B = BI = B(AA^-1) = (BA)A^-1 = IA^-1 = A^-1, and so B = A^-1

Now assume instead that AB = I. By the above result, we conclude that B is invertible and A = B^-1. But then AB = I = BA, and by Definition 5 of Matrix Operations, A is invertible and A^-1 = B.

Real Statistics Function: The following function is provided by the Real Statistics Resource Pack:

MRANK(R1, prec) = the rank of the matrix specified in range R1. The function treats numbers smaller than prec as if they were zero. If prec is omitted it defaults to .00001.

Definition 2: The range of a matrix A is the set of column vectors C for which there is a solution to the equation AX = C. The null space of A is the set of solutions to the equation AX = 0.

Observation: The range of A is simply the span of the set of columns in A. Thus the rank of A is the dimension of the range of A. A square n × n matrix is invertible if and only if the rank of A is n if and only if the dimension of the null space is 0 (i.e. the 0 vector is the only solution to the equation AX = 0).