I've been reading this paper, and I've run into an annoying issue, and I want some help verifying this is the case. In the paper, they have a matrix $A$, and they have a basis for Im$(A)$, let's say $e$. Now it seems to be the case that $A\sum e_ie_i^T = cA$ throughout this paper (for example, compare equation 27, page 8, and 40, page 11).
Note: in paper they separate $\sum_ie_ie_i^T$ into $\sum_{i=1}^se_ie_i^T+\sum_{j=1}^{\delta}f_jf_j^T$ (they use tilde e, i use f), but that is just because they correspond to different dynamics. I recombine because they are a complete basis for Im$(A)$.
Equation 27 and 40 have these $\psi$ and $\psi^{\dagger}$, with $\psi^{\dagger}\psi = (a^{\dagger},a^{2,\dagger},a^{3,\dagger})\cdot (a,a^2,a^3) = n + n^2 + n^3$, which is just a scalar (because $a^{\dagger}a = n \in \textbf{R}$), so we can commute those around. Thus when we do eq. 27 to eq. 40, we have $L = \psi A\psi^{\dagger} = \psi A(\sum_ie_ie_i^T)\psi^{\dagger}$, we can left multiply by $\psi^{\dagger}$ and right multiply by $\psi$ to get $A = A(\sum_ie_ie_i^T)$.
To understand why the equality $A = A(\sum_ie_ie_i^T)$ is true (where, again, $e_i$ are basis vectors for Im$(A)$), first notice that the normalized bases for Ker($A$) and Ker$(A)^{\perp} = $Im$(A)$ give a complete orthonormal basis. So, letting $e_i$ be basis vectors for Im$(A)$ and $f_j$ be basis vectors for Ker$(A)$, then we have that
$$A = A(\sum_{i=1}e_ie_i^T+\sum_{j=1}f_jf_j^T).$$
Yet we know that for any $v \in \textrm{Ker}(A), Av = 0$, so we have
$$A = A(\sum_{i=1}e_ie_i^T+\sum_{j=1}e_ie_i^T) = A(\sum_{i=1}e_ie_i^T)+0,$$
which is where we get $A = A(\sum_{i=1}e_ie_i^T)$, with $e_i$ being basis vectors of Im$(A)$.
The problem arises at the end, where they have worked examples. The second one, they have that (can get by computing matrix inside brackets in first two lines of eqn. 64 page 17)
$$A = \begin{bmatrix}-\alpha & \epsilon & 0 \\ \alpha & -2\epsilon & \beta \\ 0 & \epsilon & -\beta \end{bmatrix}$$
And there are two basis vectors for Im$(A)$ (these two particular vectors reached through method outlined on page 10 of paper):
$$e_1 = \begin{bmatrix} \alpha \\ 0 \\ -\beta \end{bmatrix}$$
$$e_2 = \begin{bmatrix} \alpha\beta^2 \\ -\epsilon(\alpha^2 + \beta^2) \\ \alpha^2\beta\end{bmatrix}$$
When they do $A\sum e_ie_i^T = cA$, they yield (third line of equation 64, page 17)
$$\frac{1}{\alpha^2 + \beta^2}\left(\begin{bmatrix} \alpha^2 \\ -\alpha^2 + \beta^2 \\ -\beta^2\end{bmatrix}\begin{bmatrix}\alpha & 0 & -\beta \end{bmatrix} + \begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix}\begin{bmatrix} \alpha\beta^2 & -\epsilon(\alpha^2 + \beta^2) & \alpha^2\beta \end{bmatrix}\right)$$
The fraction in the beginning comes from normalization factors (remember, we want an orthonormal basis), although there is an issue in that the square of the normalizing factor for the right-most basis vector is slightly different. And if I do $Ae_1$, I get the correct vector, $$\begin{bmatrix} -\alpha^2 \\ \alpha^2 – \beta^2 \\ \beta^2 \end{bmatrix},$$ but if I do $Ae_2$, I get
$$Ae_2 = -(\alpha^2\beta^2 + \epsilon^2(\alpha^2 + \beta^2))\begin{bmatrix}1 \\ -2 \\ 1 \end{bmatrix}$$
This is the vector wanted for $Ae_2$, but resized. This affects the weights of terms in the sum $A\sum e_ie_i^T$, and its not equal to $cA$. I'm wondering if I've missed something, or if there is an actual error here.
Edit2: going with the way they originally wrote it, you do get $A$ matrix back, just not in the way it seems it should compute to
Best Answer
Edit: (answering own question)
After figuring out the part about why $A = A(\sum_ie_ie_i^T)$ holds (see here, or brief reasoning in question that I just edited in), the rest came together. It turns out that there is no typo.
I missed something major when I didn't realize that $A = A\sum_{\alpha}g_{\alpha}g_{\alpha}^T$, for any orthonormal complete basis - the basis vectors in question must be normalized! (I am thankful to this page for giving me the push to put the pieces together, although in retrospect I feel like this should have been more obvious :| ) Since I didn't even think about this, I didn't think about normalizing the basis vectors in the problem. Let's normalize the basis vectors. To do so, we will generalize them to include a normalizing factor $c_i$, where $i \in \left\{1,2\right\}$.
$$c_1^2\begin{bmatrix}\alpha & 0 & -\beta \end{bmatrix}\begin{bmatrix} \alpha \\ 0 \\ -\beta \end{bmatrix} = c_1^2(\alpha^2+\beta^2) = 1 \implies c_1^2 = \frac{1}{\alpha^2+\beta^2}$$
$$c_2^2\begin{bmatrix}\alpha\beta^2 & -\epsilon(\alpha^2+\beta^2) & \alpha^2\beta\end{bmatrix}\begin{bmatrix}\alpha\beta^2 \\ -\epsilon(\alpha^2+\beta^2) \\ \alpha^2\beta\end{bmatrix} = c_2^2(\alpha^2\beta^4 + \epsilon^2(\alpha^2+\beta^2)^2 + \alpha^4\beta^2) = c_2^2((\alpha^2+\beta^2)((\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2))) = 1 \implies c_2^2 = \frac{1}{(\alpha^2+\beta^2)((\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2))} = \frac{1}{(\alpha^2+\beta^2)}\frac{1}{(\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2)}$$
Now, before totally diving in, notice that equation 27 and 40 tell us that $-L = \psi A\psi$, and because the calculations we want to follow in equation 64 have a positive $L$ on the left-hand side, it must be so that the right hand side must be multiplied by $-1$; we will do this at the very end. from example two of the worked examples we want to compute (using now normalized basis vectors)
$$-A(c_1^2e_1e_1^T + c_2^2e_2e_2^T).$$
Using the result from the original question and including the $c_2^2$ factor, we have for $c_2^2A\cdot e_1$:
$$c_2^2A\cdot e_1 = \frac{1}{\alpha^2+\beta^2}\begin{bmatrix}-\alpha & \epsilon & 0 \\ \alpha & -2\epsilon & \beta \\ 0 & \epsilon & -\beta \end{bmatrix}\begin{bmatrix}\alpha \\ 0 \\ -\beta \end{bmatrix} = \frac{1}{\alpha^2+\beta^2}\begin{bmatrix} -\alpha^2 \\ \alpha^2 - \beta^2 \\ \beta^2 \end{bmatrix}.$$
Now, from the original question and including the $c_2^2$ factor, we have for $c_2^2 A\cdot e_2$:
$$c_2^2A\cdot e_2 = \frac{1}{(\alpha^2+\beta^2)}\frac{1}{(\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2)}\begin{bmatrix}-\alpha & \epsilon & 0 \\ \alpha & -2\epsilon & \beta \\ 0 & \epsilon & -\beta \end{bmatrix}\begin{bmatrix}\alpha\beta^2 \\ -\epsilon(\alpha^2+\beta^2) \\ \alpha^2\beta\end{bmatrix} = \frac{1}{(\alpha^2+\beta^2)}\frac{1}{(\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2)}\begin{bmatrix}-\alpha^2\beta^2 - \epsilon^2(\alpha^2+\beta^2) \\ 2\alpha^2\beta^2 + 2\epsilon^2(\alpha^2+\beta^2) \\ -\alpha^2\beta^2 + \epsilon^2(\alpha^2+\beta^2) \end{bmatrix} = \frac{1}{(\alpha^2+\beta^2)}\frac{(\alpha^2\beta^2 + \epsilon^2(\alpha^2+\beta^2)}{(\alpha\beta)^2+\epsilon^2(\alpha^2+\beta^2)}\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix} = \frac{1}{(\alpha^2+\beta^2)}\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}.$$
So altogether, we have that
$$A(c_1^2e_1e_1^T + c_2^2e_2e_2^T) = \frac{1}{(\alpha^2+\beta^2)}(\begin{bmatrix} -\alpha^2 \\ \alpha^2 - \beta^2 \\ \beta^2 \end{bmatrix}\begin{bmatrix} \alpha & 0 & -\beta \end{bmatrix} + \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}\begin{bmatrix} \alpha\beta^2 & -\epsilon(\alpha^2+\beta^2) & \alpha^2\beta\end{bmatrix}),$$
and multiplying by $-1$ we have
$$-A(c_1^2e_1e_1^T + c_2^2e_2e_2^T) = \frac{1}{(\alpha^2+\beta^2)}(\begin{bmatrix} \alpha^2 \\ -\alpha^2 + \beta^2 \\ -\beta^2 \end{bmatrix}\begin{bmatrix} \alpha & 0 & -\beta \end{bmatrix} + \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}\begin{bmatrix} \alpha\beta^2 & -\epsilon(\alpha^2+\beta^2) & \alpha^2\beta\end{bmatrix}).$$
This is the same result as the third line of equation 64. A sigh of relief - there is no typo.