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$(\mathbf{A})<n.$ 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.


1 Answer 1


$AB$ 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 $rank(AB,C)=rank(A,C)$.

More details:

  • $rank(AB,C)$ 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 $A$.

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