Problem 6
Let A be an m × n matrix and B be an m × p matrix. Prove that
rank A B
≤ rank(A) + rank(B).
Solution
Let A ∈ Rm×n and B ∈ Rm×p where m > n and m > p, respectively.
Theorem 2.7:
Let X be an arbitrary m × n matrix. If r is the number of linearly independent rows of X and c
is the number of linearly independent columns of X, then rank(X) = r = c.
Therefore, without loss of generality let’s assume that there are rA independent columns of A, in
other words rank(A) = rA . Let those columns be denoted by A 1 , A 2 , ..., A rA . Similarly, let’s
assume that there are rB independent columns of B, in other words rank(B) = rB . Let those
columns be denoted by B 1 , B 2 , ..., B rB .
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If the matrix A is full rank, then rank(A) = rA = n. Alternatively, if the matrix A is rank-deficient,
then rank(A) = rA < n.
Similarly, if the matrix B is full rank, then rank(B) = rB = p. Alternatively, if the matrix B is
rank-deficient, then rank(B) = rB < p.
Next, let’s consider the matrix A B ∈ Rm×(n+p) .
If this matrix has full rank, then rank( A B ) = rA + rB = n + p.
However, if this matrix
is
rank deficient, then rank( A B ) is equal to the total number of independent columns of A B . By using A and B as partitioned matrices, we cannot increase the total
number of linearly independent columns by more than the number of columns of the individual
matrices A and B, respectively. Thus, this rank deficient scenario cannot exceed the value of the
full rank scenario, so we know that it must be less than the sum of the individual ranks of the two
partitioned matrices, which we denoted by rA + rB .
Therefore, we can conclude that for arbitrary matrices A and B,
rank A B
≤ rank(A) + rank(B)
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© 2022 Hayden D. Hampton, UCF