For a given positive integer \( n \), find all non-negative integers \( r \) such that the following statement holds:

For any real \( n \times n \) matrix \( A \) with rank \( r \), there exists a real \( n \times n \) matrix \( B \) such that \( \det (AB+BA) \neq 0 \).

The best solution was submitted by 홍의천 (수리과학과 2017학번). Congratulations!

Below is his solution (in text) of problem 2020-01.

An incomplete solutions was submitted by 고성훈(수리과학과 2018학번, +2).

Let’s prove the statement in the problem holds if and only if 2r>=n.
Let V=R^n and think of n*n real matrices as linear operators on V.

AB(V) and A(V), so also BA(V) has dimension at most r.
(AB+BA)(V) has dimension at most 2r so if the statement holds then 2r>=n.

Let’s assume that 2r>=n.
Let the null space and range of A be U and W, respectively, and X the intersection of the two.
Let {a_1, a_2, … , a_k}, {a_1, a_2, … , a_k, b_1, b_2, … , b_(n-r-k)},
{a_1, a_2, … , a_k, c_1, c_2, … , c_(r-k)} be bases for X, U, W, respectively.
The subspace spanned by {c_1, c_2, … , c_(n-r-k)} (here we use 2r>=n) does not contain
a non-zero element of U so {d_1, d_2, … , d_(n-r-k)}
(d_i=A(c_i), 1<=i<=n-r-k) is linearly independent.
Let {d_1, d_2, … , d_(n-r-k), e_1, e_2, … , e_(2r-n+k)} be another basis for W.
The subspace spanned by {b_1, b_2, … , b_(n-r-k)} does not contain a non-zero element of W
so {b_1, b_2, … , b_(n-r-k), d_1, d_2, … , d_(n-r-k), e_1, e_2, … , e_(2r-n+k)}
is linearly independent and we may obtain a basis for V by adding f_1, f_2, … , f_k.

Let’s define B by determining it’s values on the basis elements.
Set B(b_i) (1<=i<=n-r-k) so that AB(b_i)=c_i (which is possible since c_i is in W),
let B(d_i)=c_i+b_i (1<=i<=n-r-k) (so that AB(d_i)=d_i),
set B(e_i) (1<=i<=2r-n+k) so that AB(e_i)=e_i,
let B(f_i)=0 (1<=i<=k).
Then for any w in W, AB(w)=w.

Let’s show the range of AB+BA is the whole of V.
For 1<=i<=n-r-k, (AB+BA)(b_i)=c_i+B(0)=c_i, (AB+BA)(c_i)=c_i+B(d_i)=2c_i+b_i
so b_i is in the range of AB+BA.
For any x in X, (AB+BA)(x)=x+B(0)=x and so U is in the range of AB+BA.
For any v in V, ABA(v)=A(v) so BA(v)-v is in U.
For any w in W, (AB+BA)(w)-2w=BA(w)-w is in U so W is in the range of AB+BA.
For any v in V, (AB+BA)(v)-AB(v)-v=BA(v)-v is in U so v is in the range of AB+BA.
The whole of V is in the range of AB+BA so det(AB+BA) is not zero.