Can a nonsingular square matrix be made singular by changing exactly one element or vice versa?

Question

1. Given a nonsingular square matrix $A$, can changing just one element make it singular?
2. Given a singular square matrix $A$, can changing just one element make it nonsingular?

For $1$) I was thinking that the reduced row echelon form of $A$,say $E$ ($E$=$E_{1}E_{2}E_{3}.....E_{n}A$), must be identity so changing just one $1$ to $0$ would make it singular $\bar E$. But then , does that mean alteration in just one element in the original matrix when the same row operations are inverted on the new matrix i.e ${E_{n}}^{-1}.....{E_{1}}^{-1}\bar E = \bar A$ is singular but do $A$ and $\bar A$ differ by only one element?

For $2$) May be wrong because in this case one or more than one rows are zero rows so putting one pivotal $1$ in each row will make it non-singular and reversing the row operations will give a non-singular matrix.But the same problem here . How many of original elements are altered? If the RRE form had more than $1$ zero rows then I guess it is not just $1$ element.

Am I going in the right direction or totally messed up?

Thanks for the answers @Servaes and @Dustan Lavenstein. Those were really helpful. Can anybody please throw some light on the process I was trying ? Can the answer be obtained in that way, especially for the first question? Thanks.

Answer 1

If your base field is algebraically closed (e.g. $\Bbb{C}$) then the answer is yes:

A $k\times k$-matrix is singular if and only if its determinant is zero. The determinant is a (degree-$k$) polynomial over the base field in terms of the $k^2$ entries. Leaving one entry of the matrix variable and all others fixed therefore yields a polymial in one variable over the base field, which certainly has a root if the base field is algebraically closed. This means that for any non-singular $k\times k$-matrix, any entry can be changed to make the matrix singular. Moreover, in general there will be $k$ different values of this entry making the matrix singular.

The converse, however, is not true. If the matrix has two columns or rows that are all zeroes, then changing only one entry will leave the matrix singular. So your idea for the second question is correct. Take for example $$A=\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}.$$

Answer 2

1. Your method to get a singular matrix in the first question is not going to work. For example, you have got identity matrix as $I=EA$ where $E$ is the product of all elementary row operations. It means that $E^{-1}=A$ and after pre-multiplying with it, you basically have the identity $A=A\cdot I$. Then you switch one unity in the identity matrix to zero, but this turns the whole column in the new $A$ to zero. $$ A=\left[\matrix{a & b\\ c & d}\right], \qquad \left[\matrix{a & b\\ c & d}\right]\left[\matrix{1 & 0\\ 0 & 0}\right]=\left[\matrix{a & 0\\ c & 0}\right]. $$ A correct way to make the matrix singular by changing one element only is to look at the effect of this one element, say $x$, on the determinant. Using the cofactor expansion of the determinant along the row that contains the element $x$ we get $n$ terms, of which one is "$x$ times the cofactor of $x$" and all others do not depend on $x$, i.e. $$ \det A=ax+b $$ where $a$ is the cofactor and $b$ is a constant. If $a\ne 0$ the determinant can be set to zero by setting $x=-\frac{b}{a}$. If $a=0$ then we pick another element in the row. Clearly some element in the row must have non-zero cofactor. If all cofactors in the row are zeros then we conclude from the same expansion that $\det A=0$, which contradicts the assumption that $A$ is non-singular.
2. In your second question you may take a zero matrix as a counterexample.

P.S. As to the first question, we have proved a stronger claim. We have shown that a non-singular matrix can be made singular by changing only one element in a pre-defined row (or column).

Answer 3

For 2 ($A$ singular), the answer is YES iff $rank(A)=k-1$.

Proof. Let $A=[a_{k,l}]$; if we change $a_{i,j}$ with $a_{i,j}+x$, then $A$ becomes $A_{x,i,j}$. Then $\det(A_{x,i,j})=cofactor(a_{i,j})x+\det(A)$.

$(\Rightarrow)$ Here $\det(A)=0$ and there is $(i,j),x$ s.t. $\det(A_{x,i,j})\not= 0$. Thus $cofactor(a_{i,j})\not= 0$ and $rank(A)=k-1$.

$(\Leftarrow)$ There is $(i,j)$ s.t. $cofactor(a_{i,j})\not= 0$; thus $\det(A_{x,i,j})=cofactor(a_{i,j})x$ and we choose any $x\not= 0$.

For 1 ($A$ is invertible) The answer is YES for every $A\in GL_k$.

Proof. There is $(i,j)$ s.t. $cofactor(a_{i,j})\not= 0$. Then take $x=\dfrac{-\det(A)}{cofactor(a_{i,j})}$.