# Rowspace(B) subset of Rowspace(A) when rank(A) > rank(B)

Suppose A and B are two mxn matrices such that rank (A) > rank (B) , then I have to prove that rowspace of B is a subset of rowspace of A.

What I tried was using property of 'Simplified Span Method' , i.e. Finding reduced row echelon form after using vectors as rows of matrix, that the final rref has linearly independent non zero rows. So A will have more linearly independent vectors than B as rank(A) > rank(B). But I do not know how to proceed if these vectors are different i.e.

Suppose n is 6, so maximum possible cardinal number for a subset of mxn matrices is 6 and A contains 3 of these and B contains 2 and the other one is left out. In this case, rowspace of B will not be a subset of rowspace of A.

• It's not hard to find two $3\times3$ matrices $A,B$ such that the rank of $A$ is 2, the rank of $B$ is 1, but the rowspace of $B$ is not a subset of the rowspace of $A$. – Gerry Myerson Feb 19 '17 at 10:27
• Did you mean in your question to ask how to disprove? – Maciej Caputa Feb 19 '17 at 10:39
• The question asked to prove or disprove the fact, I completely ignored that. 😅 – Quark Feb 19 '17 at 10:41

This statement is not correct and can be disproved by a counter example as you mentioned.

Let $A$ and $B$ be a $3$ by $3$ matrices with $$Rowspace(A) = span\left\{<1,0,0>,<0,1,0>\right\}$$ and $$Rowspace(B) = span\left\{<0,0,1>\right\}$$

Hence $Rank(A) > Rank(B)$, but $Rowspace(B) \not\subset Rowspace(A)$

However, following statement is true.

$$A, B \in M_{m,n}(\mathbb{R}) : rank(A) = m\ \wedge\ rank(A) > rank(B) \implies Rowspace(B) \subseteq Rowspace(A)$$

This can be read as:

Let $A$ and $B$ be m by n matrices with real entries. If rowspace of A spans $\mathbb{R}^m$ ($dim(Rowspace(A)) = m$) and $rank(A) > rank(B)$ then rowspace of B is a proper subset of rowspace of A.