Determinant of a sum of two matrices For a matrix $X$ with $\det X = 0$, what should be the constraints on $Y$ such that $\det (X+Y) = 0$? On obvious choice would be $Y=\alpha X$, but can one say something more than this?
 A: In general there is no good way to describe such $Y$. Already for $n=2$ we can see how the situation is like. Take for example $X=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}$ and compute all possible $Y$ with $\det(X+Y)=0$.
One can, however, consider special cases. For example, one can use the Matrix Determinant Lemma, which says the following:
$$\det(X+Y)=\det(uv^T+Y)=(1+v^TY^{-1}u)\det(Y),$$
where $Y$ is an invertible matrix and $v^TY^{-1}u$ is interpreted as a scalar. Then, with $X=uv^T$, we always have  $\det(X)=0$. If $v^TY^{-1}u=1$, then $\det(X+Y)=0$.
A: Let $X+Y=[(x_1+y_1)\,\,(x_2+y_2)\,\,\dots (x_n+y_n)]$, namely $x_i+y_i$ represents the $i$-th column of your matrix. Clearly, $det(X+Y)=0$, iff $X+Y$ is singular, that is, there exists (at least) a column, viz. $k$, that is linear combination of the others:
$a_k+b_k=\sum_{i=1\\i \neq k}^{n}c_i(a_i+b_i)$ (Eq. 1)
Let $e_k=[0\dots0\,1\,0\dots]^T$ be the vector having all-zero entries, except the $k$-th, which is unitary. Therefore:
$a_k+b_k=(X+Y)e_k$ (Eq. 2)
By analogy, let $C=[c_1\,c_2\dots 0 \dots c_n]^T$ be a vector having arbitrary real values $c_i$, except the $k$-th, which is 0. Hence:
$\sum_{i=1\\i \neq k}^{n}c_i(a_i+b_i)=(X+Y)C$ (Eq. 3)
Replacing Eqs. (2), (3) into (1) yields:
$(X+Y)e_k=(X+Y)C$ (Eq. 4)
Eq. (4) can be expressed as a linear system by using the Kronecker Product and vectorising Eq. (4) itself, namely:
$[(e_k^T\bigotimes I_n)-(C^T\bigotimes I_n)]vec(X+Y)=0$ (Eq. 5)
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Eq. (5) shows that $X+Y$ be singular can be turned into finding the null-space of the set of matrices $(e_k^T\bigotimes I_n)-(C^T\bigotimes I_n),\qquad \forall k\in [1,n],\, C \in R^n, s.t. C_k=0$.
A similar argument applies to the matrix $X$ that, in turn, is singular. Hence:
$[(e_j^T\bigotimes I_n)-(D^T\bigotimes I_n)]vec(X)=0,\forall j\in [1,n],\, D \in R^n, s.t. D_j=0$ (Eq. 6)
Both Eqs. (5) and (6) give the general constraints to your problem. The formulation may be quite involved, but still linear. So, I think that finding the space of solutions won't be that hard by using a software able to deal with linear algebra!
A: If $X + Y = V$ is a Vandermonde matrix with $V_{i,j} = v_i^{j-1}$ and the $v_i$ are not distinct then $|V| = 0$.
These are one family of solutions and there are others.
