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, …, Cp be any basis for the span of the column vectors in A. Thus any column Aj in A can be expressed uniquely as a linear combination of the basis elements, and so there are bij such that Ai = \sum_{k=1}^p b_{kj} C_k. Thus A = CB where C is the m × p matrix whose columns are C1, …, Cp and = [bij], a p × n matrix.

Since A = CB, every row in A can be expressed as a linear combination of the rows in B, namely \sum_{k=1}^m c_{ik} B_k. This means that 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 AT, we conclude that rank(AT ) ≤ column rank of AT. But clearly rank(AT ) = column rank of A and column rank of AT = rank(A). Thus we conclude that column rank of A ≤ rank(A). Putting both conclusion 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, = 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 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).

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