Let A be a m x n complex matrix with rank A = r . If we are allowed to modify at most one row ( or at most one column , but not both ) of A , then what effect will it have on rank A ?
Thanks for any help .
Let A be a m x n complex matrix with rank A = r . If we are allowed to modify at most one row ( or at most one column , but not both ) of A , then what effect will it have on rank A ?
Thanks for any help .
Let $\tilde{A} \in \mathbb{C}^{m \times n}$ be the matrix obtained by changing a row (or) column of $A \in \mathbb{C}^{m \times n}$. Then the claim is that $-1 \leq \text{rank}(A) - \text{rank}(\tilde{A}) \leq 1$
Proof:
We will assume that the $k^{th}$ row of $A \in \mathbb{C}^{m \times n}$ has been modified to get $\tilde{A} \in \mathbb{C}^{m \times n}$. Then note that $\tilde{A} = A + \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0\\ r_{k,1} & r_{k,2} & r_{k,3} &\cdots & r_{k,n-1} & r_{k,n} \\ 0 & 0 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0\end{bmatrix}_{m \times n}$ where $r_{k,\ell} \in \mathbb{C}$. Note that this can be written as
$\tilde{A} = A + \begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}_{m \times 1}\begin{bmatrix} r_{k,1} & r_{k,2} & r_{k,3} &\cdots & r_{k,n-1} & r_{k,n}\end{bmatrix}_{1 \times n}$ Hence, it is equivalent to adding a rank $1$ perturbation to the matrix $A$. Hence, $\text{rank}(\tilde{A}) \in \{\text{rank}(A)-1,\text{rank}(A), \text{rank}(A)+1 \}$
The rank of a matrix is equal to both the number of linearly independent rows and the number of linearly independent columns. Therefore, if you change at most one row or column, the rank can either increase by one, decrease by one, or stay the same. However, you can't say much more than this since you don't know anything about the structure of the matrix or how you're making the change.