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Let $\xi$ and $\zeta$ be two covectors in $V^*$. What can we say about the $(1,1)$ tensor $$ \xi \otimes \zeta^\sharp + \text{tr}(\xi\otimes \zeta^\sharp)I? $$ Here $\zeta^\sharp$ is the metric dual to $\zeta$ (assume we have some inner product - which is Lorentzian in my case - so that this is defined). More concretely I am wondering if the resulting matrix representation is a nonsingular matrix. The context is that I am trying to prove existence and uniqueness of solutions to the system of equations $$ (a b^T + b^T a I)x = c, $$ where $a, b, c$ are known vectors. Here $I$ is the $n \times n$ identity matrix.

More context/conditions: The covectors $\xi, \zeta$ are both null with respect to an ambient Lorentzian inner product $g$ on $V$. Furthermore, $g(\xi, \zeta) = -2$. Therefore the covectors are linearly independent.

Chris
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  • For example, with $\xi=(1,0,0,...)$ and $\zeta$ the "transpose/dual" of that, the trace is $1$, and the rank-one tensor $\xi\otimes\zeta^\sharp$ can be put into coordinates as a matrix with all $0$'s except a solitary $1$ in the upper left corner. So your expression would be the identity matrix with one of the diagonal $1$'s converted to $0$. Not invertible. – paul garrett Oct 01 '22 at 19:42
  • And, really, trace of $\xi\otimes \zeta^\sharp$ is $\zeta^\sharp(\xi)$, while the rank-one operator/tensor $\xi\otimes \zeta^\sharp$ has just one non-zero eigenvalue, namely, that trace, etc. So the same pattern occurs generally... – paul garrett Oct 01 '22 at 19:44
  • Thanks for the example and comment, these are super helpful. I forgot to add the conditions that prevent this from occurring, since I hadn't thought of an example like this. I'll add them now. – Chris Oct 01 '22 at 20:52
  • related? https://math.stackexchange.com/questions/219731/determinant-of-rank-one-perturbations-of-invertible-matrices (this may be helpful to study the determinant of $ab^t+b^ta I$) – Giulio R Oct 01 '22 at 21:04

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