A sequence of square matrices with lim (det) = 0 but condition number of 1

Question

When the condition number of a matrix is $1$, it implies it is invertible (nonsingular) and hence its determinant is nonzero. Is this also true in the limit? Specifically, is there a sequence of square matrices with $$ \lim_{n\to\infty} \det(M_n)=0$$ but with $\kappa_2(M_n)=1$ for all $n$?

Answer 1

Yes, there is such a sequence. The condition number of the identity matrix $I$ is $1$, as is its determinant. Let $$M_n = \frac{1}{n}I.$$ Then $\det(M_n) = (1/n)^d$ where $d$ is the dimension. But $\kappa_2(M_n) = 1$ for all $n$, so this is an example of such a sequence.