# Question about the rank of an augumented matrix

Suppose that we have three matrices: $$\mathbf{A, \, B}$$ are of size $$n\times n,$$ and $$\mathbf{C}$$ is of size $$n\times m.$$ I wanted to know that, if $$\mathbf{B}$$ is nonsingular, then rank$$(\mathbf{AB}, \mathbf{C})=$$ rank$$(\mathbf{A, C})?$$ Here comma stands for the augumented matrix.

Here is how I am proceeding. First of all we note that rank$$(\mathbf{AB})=$$ rank$$(\mathbf{A}),$$ if $$\mathbf{B}$$ is invertible.

I will split the possibilities into two cases.

Case (i): If $$\mathbf{A}$$ is also nonsingular, then $$\mathbf{AB}$$ will be nonsingular, hence both will have a full rank $$n.$$ Therefore

rank$$(\mathbf{AB}, \mathbf{C})=$$rank$$(\mathbf{A, C}).$$

Case(ii): Let $$\mathbf{A}$$ be singular, then $$\mathbf{AB}$$ is also singular. Further rank$$(\mathbf{AB})=$$ rank$$(\mathbf{A}) Now everything will be decided by the matrix $$\mathbf{C}.$$

Suppose for example if I have a matrix $$\mathbf{A}=\begin{pmatrix}0 & 0\\ 1 & 0 \end{pmatrix}$$ and $$\mathbf{AB}=\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix},$$ for some invertible matrix $$\mathbf{B},$$ then if we have $$\mathbf{C}=\begin{pmatrix} 1 \\ 0 \end{pmatrix},$$ then my answer is negative to the question. But I wanted to know whether this situation is possible. More explanation is needed. And also if the answer is yes, then what is the general proof.

$$AB$$ and $$A$$ share the same column space. Hence the example that you illustrated is not possible.
Since $$AB$$ and $$A$$ share the same column space, we have $$rank(AB,C)=rank(A,C)$$.
• $$rank(AB,C)$$ is equal to the number of linearly independent columns, in $$[AB, C]$$. We first get a basis for the column of $$AB$$ then check if there are any vectors that are not spanned by the the basis of $$AB$$ and add those vector to it to count the rank. We can use the same basis for column space of $$AB$$ for the column space of $$A$$.