Relation between Eigenvalues of a given matrix and another set of rank-one matrices. Let $\mathbf{T}=[\mathbf{t}_1,\dots,\mathbf{t}_d]$ be a $m\times d$ matrix with $\mathbf{t}_i$ as its linearly independent columns. Also I assume $d<\min(m,n)$. Let $\mathbf{H}$ be a $n\times m$ matrix. Let $\mathbf{W}$ be a $n \times n$ positive definite matrix. For $i=1,\dots,d$, let me define the matrices
\begin{align}
\mathbf{B}_i&=\mathbf{H}\mathbf{t}_i\mathbf{t}_i^H\mathbf{H}^H \\
\mathbf{C}_i&=\mathbf{W}+\sum_{k\neq i}\mathbf{H}\mathbf{t}_k\mathbf{t}_k^H\mathbf{H}^H \\
\mathbf{D}_i&=\mathbf{C}_i^{-1/2}\mathbf{B}_i\mathbf{C}_i^{-1/2}
\end{align}
Here $\mathbf{A}^H$, $\mathbf{A}^{-1}$,$\mathbf{A}^{1/2}$ denotes hermitian, inverse and cholesky root (or square root)  of matrix $\mathbf{A}$ respectively. 
Now all $\mathbf{D}_i$'s are rank-one matrices since each $\mathbf{B}_i$ is also a rank-one matrix. So they have one non-zero eigenvalue each. Let the non-zero eigenvalue of  $\mathbf{D}_i$ be $\alpha_i$ for all $i\in\{1,\dots,d\}$
CLAIM: $\alpha_1,\dots,\alpha_d$ are also the eigenvalues of $\mathbf{T}^H\mathbf{H}^H\mathbf{W}^{-1}\mathbf{H}\mathbf{T}$
Is it true. If so, how do I prove this?. It is becoming really difficult for me. 
 A: Let $u_i=W^{-1/2}Ht_i$ and $U=\pmatrix{u_1&u_2&\cdots&u_d}\in M_{n\times d}(\mathbb{C})$. Then $\alpha_i=\|C_i^{-1/2}Ht_i\|^2=u_i^H(W+UU^H - u_iu_i^H)^{-1}u_i$ and $T^HH^HW^{-1}HT=U^HU$. In this formulation, since $U^HU$ depends solely on $U$ but $\alpha_i$ depends on both $U$ and $W$, there is no reason to believe that $\alpha_i$ is an eigenvalue of $U^HU$.
A: For the matrices of interest I can show a shared null space of their similar matrices. This may not be helpful but I hope it is, and I did the work so I thought I would show it.
Let $\mathbf{M} = \mathbf{H}\mathbf{T}$. In matrix notation, using $\mathbf{e}_0^H = \pmatrix{1 & 0 & 0 & \cdots}$, your equations are
$$\mathbf{B}_i = \mathbf{M}\mathbf{e}_i\mathbf{e}_i^H\mathbf{M}^H$$
and
$$\mathbf{C}_i = \mathbf{W} +\mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H$$
The first matrix has similarity
$$\mathbf{D}_i \sim \mathbf{B}_i\mathbf{C}_i^{-1}\tag{1}$$
The second matrix has similarity
$$\mathbf{M}^H\mathbf{W}^{-1}\mathbf{M}\sim \mathbf{M}\mathbf{M}^H\mathbf{W}^{-1}\tag{2}$$
Let $\mathbf{V}$ be the right pseudo-inverse of $\mathbf{M}^H$ so that $\mathbf{M}^H\mathbf{V} = \mathbf{I}$. The two matrices have the same right null space
$$\left(\mathbf{B}_i\mathbf{C}_i^{-1}\right)\left[\mathbf{W}\mathbf{V}\mathbf{M}^H-\mathbf{W}\right] = 0 \tag{A}$$
and
$$\left(\mathbf{M}\mathbf{M}^H\mathbf{W}^{-1}\right)\left[\mathbf{W}\mathbf{V}\mathbf{M}^H-\mathbf{W}\right] = 0 \tag{B}$$
which is a little surprising to me given the way $\mathbf{C}_i$ is built using $\mathbf{W}$ and that (A) is true for each $i$. (B) is relatively obvious, so I will show (A). (A) is true if and only if
$$\mathbf{B}_i\mathbf{C}_i^{-1}\mathbf{W}\mathbf{V}\mathbf{M}^H = \mathbf{B}_i\mathbf{C}_i^{-1}\mathbf{W} $$
Using from the definition of $\mathbf{C}_i$ that
$$\mathbf{W} = \mathbf{C}_i - \mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H \tag{W}$$
we have
$$\mathbf{C}_i^{-1}\mathbf{W} = \mathbf{C}_i^{-1}\left[\mathbf{C}_i - \mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H\right]$$
$$= \mathbf{I} - \mathbf{C}_i^{-1}\mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H$$
and from $\mathbf{V}$ a right inverse of $\mathbf{M}^H$
$$\mathbf{C}_i^{-1}\mathbf{W}\mathbf{V}\mathbf{M}^H = \mathbf{V}\mathbf{M}^H - \mathbf{C}_i^{-1}\mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H$$
and therefore
$$\mathbf{B}_i\mathbf{C}_i^{-1}\mathbf{W}\mathbf{V}\mathbf{M}^H = \underbrace{\overbrace{\mathbf{B}_i\mathbf{V}}^{\mathbf{M}\mathbf{e}_i\mathbf{e}_i^H}\mathbf{M}^H}_{\mathbf{B}_i} - \mathbf{B}_i\mathbf{C}_i^{-1}\mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H$$
$$=\mathbf{B}_i\left[\mathbf{I} - \mathbf{C}_i^{-1}\mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H\right]$$
$$=\mathbf{B}_i\mathbf{C}_i^{-1}\left[\mathbf{C}_i - \mathbf{M}\left(\mathbf{I} - \mathbf{e}_i\mathbf{e}_i^H\right)\mathbf{M}^H\right]$$
$$=\mathbf{B}_i\mathbf{C}_i^{-1}\mathbf{W}$$
where the last step uses the equation (W) for the equality on $\mathbf{W}$
