I was just thinking about this problem:
Can every nonsingular $n\times n$ matrix with real entries be made singular by changing exactly one entry?
Thanks for helping me.
Linear Algebra – Can Every Nonsingular n×n Matrix be Made Singular by Changing One Entry?
linear algebramatrices
Best Answer
If $A$ is a nonsingular matrix with rows $r_1,r_2,\ldots,r_n$, then $\{r_2,\ldots,r_n\}$ spans an $(n-1)$-dimensional subspace $P$ of $\mathbb R^n$. At least one of the standard basis vectors $e_1,e_2,\ldots,e_n$ is not in $P$, say $e_i$. Then $\{e_i,r_2,r_3,\ldots,r_n\}$ is a basis of $\mathbb R^n$, and it follows that there is a real number $c$ such that $r_1-ce_i$ is in $P$. The matrix $A'$ with rows $(r_1-ce_i),r_2,r_3,\ldots,r_n$ is singular, and it is obtained from $A$ by subtracting $c$ from the entry in the first row and $i^\text{th}$ column.
Here's a way to rephrase this somewhat more geometrically. The subspace $P$ is a hyperplane that divides $\mathbb R^n$ into two half-spaces, and $r_1$ lies in one of these halves. The line through $r_1$ in the direction of a vector $v$ has the form $\{r_1+tv:t\in\mathbb R\}$. This line is parallel to $P$ only if $v$ is in $P$; otherwise, the line will cross $P$. Since $P$ can't be parallel to all of the coordinate directions (or else it would fill up all of $\mathbb R^n$), there must be a line of the form $\{r_1+te_i:t\in\mathbb R\}$ that crosses $P$, where $e_i$ is the standard basis vector with a $1$ in the $i^\text{th}$ position and $0$s elsewhere. This means that there exists $t_0\in \mathbb R$ such that $r_1+t_0e_i\in P$. And then, linear dependence of the vectors $r_1+t_0e_i,r_2,\ldots,r_n$ means that the matrix with those rows is singular.