Solving a tensor equation from singularity condition I apologize if this question has already been asked but I'm not sure what the best key-words are.
I have a tensor equation of the form: $(A + x\otimes b)c=0$.
Here, $A$ is a second-order tensor/matrix, and the rest are first-order tensors/vectors. $A, b$ are known but $x, c$ are unknown. I am interested in calculating a vector '$x$' such that $A + x\otimes b$ is singular. What would be an approach to solve such a problem?
Thanks!
 A: If $A$ is a square matrix, then it has a Jordan canonical form:
\begin{equation}
P^{-1}AP = J = \left(\begin{array}{ccc}
J_{i_1}(\lambda_1) & & \\
& \ddots & \\
& & J_{i_n}(\lambda_n)
\end{array}\right),
\end{equation}
where

*

*$\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$

*each $J_{i_m}(\lambda_m)$ is an $i_m\times i_m$ Jordan block.

For example, a $2\times 2$ Jordan block has the form
\begin{equation}
J_2(\lambda) = \left(\begin{array}{c}\lambda & 1\\0 & \lambda\end{array}\right).
\end{equation}
Let's suppose that $A$ is equal to a Jordan form. We address the other case afterward.
When dealing with concrete matrices, I prefer the notation $xb^{\mathsf{T}}$ over $x\otimes b$. If $x$ and $b$ are column vectors, then $xb^{\mathsf{T}}$ is a rank-1 matrix; each column is a scalar multiple of any other column in this matrix.

We assume that $A$ is in Jordan canonical form.
If the top entry of $b$ is not $0$, then we can engineer an $x$ so that the first column of $A - xb^{\mathsf{T}}$ is all zeros. Since $A$ is Jordan form, it has the form
\begin{equation}
A =
\left(\begin{array}{c}
\lambda_1 & \cdots\\
0 & \ddots \\
\vdots & \ddots\\
0 & \cdots
\end{array}\right),
\end{equation}
i.e. its first column has an eigenvalue in its top entry and zeros otherwise.
Consider
\begin{equation}
x = \left(\begin{array}{c}\lambda_1/b_1\\0\\\vdots\\0\end{array}\right).
\end{equation}
If we set $x$ to this, then
\begin{equation}
xb^{\mathsf{T}} =
\left(\begin{array}{c}
\lambda_1/b_1\\
0\\
\vdots\\
0
\end{array}\right)
\begin{array}{c}
\left(b_1, b_2,\ldots,b_N\right)\\
\\
\\
\\
\end{array}
=
\left(\begin{array}{ccc}
\lambda_1 & \lambda_1\cdot b_2/b_1 & \cdots\\
0 & 0 & \cdots\\
\vdots & \vdots & \ddots\\
0 & 0 & \cdots
\end{array}\right).
\end{equation}
With this engineered $xb^{\mathsf{T}}$, the first column  of $A - xb^{\mathsf{T}}$ is all zeros, and $A - xb^{\mathsf{T}}$ is singular.

Now suppose that $A$ is not in Jordan canonical form, but $J = P^{-1}AP$ is. If $A - xb^{\mathsf{T}}$ is singular, then $P^{-1}\left(A - xb^{\mathsf{T}}\right)P$ is, too.
\begin{eqnarray}
P^{-1}\left(A - xb^{\mathsf{T}}\right)P &=& P^{-1}AP - P^{-1}xb^{\mathsf{T}}P\\
&=& J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}
\end{eqnarray}
We consider the same scheme as before. This time, however, we want $P^{-1}x$ to have the form
\begin{equation}
P^{-1}x =
\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right).
\end{equation}
To ensure that this happens, set
\begin{equation}
x =
P\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right),
\end{equation}
so that
\begin{equation}
P^{-1}x ~=~ 
P^{-1}P\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right)
~=~
\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right).
\end{equation}
With $x$ engineered this way, we have
\begin{equation}
\left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}
=
\left(\begin{array}{c}
\lambda_1/b_1\\
0\\
\vdots\\
0
\end{array}\right)
\begin{array}{c}
\left((P^{\mathsf{T}}b)_1, (P^{\mathsf{T}}b)_2,\ldots,(P^{\mathsf{T}}b)_N\right)\\
\\
\\
\\
\end{array}
=
\left(\begin{array}{ccc}
\lambda_1 & \lambda_1\cdot (P^{\mathsf{T}}b)_2/(P^{\mathsf{T}}b)_1 & \cdots\\
0 & 0 & \cdots\\
\vdots & \vdots & \ddots\\
0 & 0 & \cdots
\end{array}\right).
\end{equation}
This ensures that the first column of $J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}$ is all zeros. Thus $J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}$ is singular, and $A - xb^{\mathsf{T}}$ is, too.
