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.

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