Norm of diference of matrices of different rank

Question

Suppose $A$ is a $n\times n$ matrix of rank $k$ that has Euclidean norm equal to $1$. Given $p<k$, and $\epsilon>0$, can we always find a norm one matrix $B$ of rank $p$ such that $||A-B||<\epsilon$?

Edit: In view of the answer below, what if we require $k<n$?

Answer 1

This is in fact never true; suppose you have a matrix $A$ of rank $k$. Then, we can take linearly independent $v_1, \dots, v_k \in \text{Ran}(A)$, so there exist $u_1, \dots, u_k$ such that $Au_i = v_i$ for each $i$. Suppose for a contradiction the statement is true for some $p < k$, so there is a sequence $(B_n)$ of $n \times n$ norm 1 matrices of rank $p$ which converges to $A$. Then, for each $i$, $\lim_{n \to \infty} B_nu_i = Au_i = v_i$. If we take $n$ large enough, each $B_nu_i$ will be close enough to $v_i$ that $\{B_nu_1, \dots, B_nu_k\}$ is linearly independent, so $B_n$ will have rank at least $k>p$, yielding a contradiction.

Answer 2

No. The determinant is a continuous map on $\mathbb{R}^{n\times n}$. If $A$ is a matrix with nonzero determinant (full rank), then any matrix $B$ sufficiently close to $A$ will also have nonzero determinant (and thus will also be full rank), by continuity.