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For $1 \leq i \leq n$, let $A=\begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \\ \end{bmatrix}$, $B_i=\begin{bmatrix} b_{i1} \\ \vdots \\ b_{in} \end{bmatrix}$ and $C_i=\begin{bmatrix} c_{i1} \\ \vdots \\ c_{in} \end{bmatrix}^*$.

Let $D=A+\sum_{1 \leq i \leq n}B_i k_i C_i$. Then, for almost all $a_{ij}$, $b_{ij}$, $c_{ij}$, there exists $k_i \in \mathbb{C}$ such that all eigenvalues of $D$ are zeros.

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    $\begingroup$ Why do you think this is true? Does the $2\times 2$ case work? $\endgroup$
    – Yemon Choi
    Jan 26 '12 at 7:52
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    $\begingroup$ Consider the characteristic polynomial $P(k_1,\ldots,k_n,\lambda) = \det(D - \lambda)$. For all eigenvalues of $D$ to be $0$, the coefficients of $\lambda^0$ to $\lambda^{n-1}$ must be $0$. That makes $n$ polynomial equations in the $n$ variables $k_1, \ldots, k_n$. I would expect that for almost every choice of the $a_{ij}, b_{ij}, c_{ij}$ this system of equations will have at least one solution in ${\mathbb C}^n$. $\endgroup$ Jan 26 '12 at 8:33
  • $\begingroup$ Consider the matrix $C$ made by stacking the $C_i$ together, the matrix $K$ with $k_i$ on the diagonal and $0$ elsewhere, and the matrix $B$ by lining up the $b_i$ against each other. We can then write: $D=A+BKC$. In general position, $C$ is invertible, so we can conjugate by it and change $D$ and $B$ as appropriate, and we are left with $D=A+BK$, with $A$ and $B$ generic and $K$ chosen from among diagonal matrices. I think we want the eigenvalues of $D$ to all be different. If those are different and $B$ and $C$ are invertible, I think it should work. $\endgroup$
    – Will Sawin
    Jan 26 '12 at 21:46
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    $\begingroup$ Does anyone know what the tangent space of the variety of nilpotent matrices looks like? If it did not contain any invertible matrices, or matrices in any other "generic" class, that would solve the problem by a dimension argument. $\endgroup$
    – Will Sawin
    Jan 26 '12 at 22:00
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The variety $N$ of nilpotent matrices has codimension $n$ (e.g., this MSE answer). In projective space, generally, $A,B_1C_1,B_2C_2,\dotsc,B_nC_n$ span an $n$-plane. Because of the dimension, this plane meets $N$. So there is some combination $k_0 A + k_1 B_1 C_1 + \dotsb + k_n B_n C_n$ which is nilpotent, with not all of the $k_i$ equal to zero.

Generally an $(n-1)$-plane misses $N$. If the plane spanned by the $B_i C_i$ (without $A$) misses $N$, then it forces $k_0$ in the above linear combination to be nonzero. Dividing by $k_0$ gives $A + k'_1 B_1 C_1 + \dotsb + k'_n B_n C_n$ nilpotent.

So, does the $(n-1)$-plane spanned by the $B_iC_i$ miss $N$? It's not quite a general $(n-1)$-plane because it's spanned by rank one matrices. But it still misses $N$ anyway. The family of $(n-1)$-planes spanned by rank one matrices is irreducible, so it's sufficient to show a single one that misses $N$. Take the rank one matrices with a $1$ in one position on the diagonal and all other entries zero. These are rank one, linearly independent, and no linear combination of them is nilpotent.

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[update] By an example of 4x4-matrices the ansatz below could not be used to solve the problem. The matrix $\small Q_K $ cannot in general be made lower triangular by choices of the $\small k_i $. I'll delete this answer soon if I cannot improve the ansatz.


I do not yet have a full answer but possibly a first step into one. I think that this can be answered unsing two facts:

a) There is a similarity transformation with a rotation T such that $\small P = T^{-1}\cdot A \cdot T = T' \cdot A \cdot T $ where P is triangular and has the eigenvalues of A on its diagonal.

b) Your sum-expression of vector outer products, let it be called matrix $\small E_K = \sum_{i=1}^n B_i k_i C_i = \sum_{i=1}^n k_i (B_i C_i)= \sum_{i=1}^n k_i E_i $ is a weighted (by the $\small k_i $ weights) sum of rank-1-matrices $\small E_i $

From the "similarity rotated version" of all matrices

$\qquad \small Q_K = T' E_K T = \sum_{i=1}^n k_i (T' E_i T) = \sum_{i=1}^n k_i Q_i $ (which should be made triangular by choices of $\small k_i $ ) and
$\qquad \small R = T' D T $ which is then also triangular

we get your final equation in its form with triangular matrices

$\qquad \small R = P + Q_K $

We'll have a solution if the weights $\small k_i $ for the non-triangular, generic but rank-1-matrices $\small Q_i $ can be chosen such that their sum $\small Q_K$ becomes triangular and its diagonal equals the negative diagonal in $\small P $.

I've a vague speculation that the equation with this triangular matrices can easier be shown to be "almost always" possible, but have not yet a further concrete approach. At least this construction exhibits that the rank-1-matrices $\small E_i $ (and thus $\small Q_i$ ) cannot be simple scalar multiples of each other if A has full rank and thus the restriction to "for almost all" is unavoidable and possibly mainly consists of this property.

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  • $\begingroup$ I think this method doesn't work. You can't find the common $T$ such that $T^{-1} A T$ is triangular and $T^{-1} Q_K T$ is also triangular. $\endgroup$
    – Seyong
    Jan 27 '12 at 3:41
  • $\begingroup$ @Seyong: seems to be true. I tried with a 4x4-example and the $\small k_i $ could not be chosen such that the strict upper triangle of $\small Q_K$ is zero: I need one of the $\small k_i$-parameters to get one entry in $\small Q_K$ zero but we have $\small n (n-1)/2 $ such entries to be made zero simultaneously according to my ansatz. I hoped, that the rank-1-property of the $\small Q_i $-matrices could be exploited, but that seems not be sufficient this way. $\endgroup$ Jan 27 '12 at 7:12

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