Question

Consider an n x p matrix A and a p x m matrix B.

a. What can you say about the relationship between rank(A) and rank(AB)?

b. What can you say about the relationship between rank(B) and rank(AB)?

Short answer

a. The relationship between rank(A) and rank(AB) is

$rank\left(AB\right)\le rank\left(A\right)$

b. The relationship between rank(A) and rank(AB) is

$rank\left(AB\right)\le rank\left(B\right)$

Step-by-step solution

(a) Considering the matrix A

Let $r_1\mathrm{and}{r}_2$be the rank of the matrices A and B.

$\because r_1$is the rank of the matrix A

$\therefore A~{\left[\begin{array}{c}M\\ 0\end{array}\right]}_{nxp}$

Where M is a sub-matrix of rank$r_1$ and contains$r_1$ rows.

  Finding the rank of the matrix AB

$\therefore A~{\left[\begin{array}{c}M\\ 0\end{array}\right]}_{nxp}$

Post-multiplying it by matrix B, we get

$AB~\left[\begin{array}{c}M\\ 0\end{array}\right]B$

But $\left[\begin{array}{c}M\\ 0\end{array}\right]B$can have $r_1$non-zero rows at most which are obtained on multiplying $r_1$non-zero rows of sub-matrix M with column of B.

$$\begin{gathered} \Rightarrow rank\left(AB\right)=rank\left\{\left[\begin{array}{c}M\\ 0\end{array}\right]B\right\}\le r_1 \\ \Rightarrow rank\left(AB\right)\le r_1 \\ \Rightarrow rank\left(AB\right)\le rank\left(A\right) \end{gathered}$$

(b) Considering the matrix B

Let $r_1\mathrm{and}{r}_2$be the rank of the matrices A and B.

$\because r_2$is the rank of the matrix B.

$\therefore B~[N0]$

Where N is a sub-matrix of rank$r_2$ and contains$r_2$ columns.

Finding the rank of the matrix AB

$\because B~\left[\right(N0]$

Pre-multiplying it by matrix A, we get

localid="1659357430337" $AB~A\left[\begin{array}{cc}N& 0\end{array}\right]$

But $A\left[N0\right]$can have $r_2$non-zero columns at most which are obtained on multiplying $r_2$non-zero columns of sub-matrix N with rows of A.

$$\begin{gathered} \Rightarrow rank\left(AB\right)=rank\left\{\right[(N0)\left]\right\}\le r_2 \\ \Rightarrow rank\left(AB\right)\le r_2 \\ \Rightarrow rank\left(AB\right)\le rank\left(B\right) \end{gathered}$$

  Final Answer

Since, the matrix A is of order n x p and matrix B is of order p x m then

a. The relationship between rank(A) and rank(AB) is

$rank\left(AB\right)\le rank\left(A\right)$

b. The relationship between rank(A) and rank(AB) is

$rank\left(AB\right)\le rank\left(B\right)$