For every matrix $A$, can one find a matrix $B \ne O$ such that $\det(A+B)=\det(A)+\det(B)$?

Question

Since we can't say that $\det(A+B)=\det(A)+\det(B)$ for every matrix $A$ and $B$, we can ask a different question. Clearly, for every matrix $A$ we have $\det(A+O)=\det(A)+\det(O)$. So can we find for every matrix $A$ a matrix $B\neq O$ such that $\det(A+B)=\det(A)+\det(B)$?

If $A$ is singular we get $\det(A-A)=\det(A)+\det(-A)$ and we can find many other ways to obtain our desired result. How do we deal with the case of $A$ being invertible?

Answer 1

If $A$ is arbitrary, invertible or not, let $B$ be the matrix obtained by interchanging the first two columns of $A$.

Answer 2

It can always be done, in a rather trivial way.

Suppose first that $A$ is in Jordan form. Then $\det A$ is the product of its diagonal entries. Let $B=E_{12}$, the matrix with a $1$ in the 1,2 entry and zeroes elsewhere. Then $\det B=0$, and $\det(A+B)=\det A$ (since $A+B$ is upper triangular), so $$\det(A+B)=\det A+\det B.$$

For general $A$, write $A=SJS^{-1}$ with $J$ its Jordan form. Let $B=SE_{12}S^{-1}$. Then $$ \det(A+B)=\det(S(J+E_{12})S^{-1})=\det (J+E_{12})=\det J=\det A=\det A+\det B. $$

Answer 3

Let $H$ be a nonzero triangular matrix with zero's on the diagonal (assuming the dimension of the matrices is $\ge2$), and let $$B:=AH.$$ Then, $B\ne O$ if $A$ is regular, and $$\det(I+H)=1=\det(I)+\det(H)$$ hence (multiplying on the left by $\det A$) $$\det(A+B)=\det A+\det B.$$

Answer 4

There are three possibilities:

1. If $A=0$ or $A$ is $1\times1$, any nonzero $B$ will do.
2. If $A$ is nonzero and singular, pick $B=A$.
3. If $A$ is nonsingular and it is at least $2\times2$, pick $B=AN$, where $N$ is any nonzero nilpotent matrix.

Answer 5

Consider the equation in $B$ $$f(B) \colon = \det(A+B) - \det A - \det B = 0$$

At $B = 0$ the gradient of $f$ equals the cofactor matrix of $A$. If this equals the $0$ matrix, then also $\det A=0$ ( and we can take $B=A$ as a solution of the problem). Otherwise, there exists a full manifold of codim $1$ of solutions around $B=0$.