Let $x_t \in \mathbb{S}^{d-1}$, $\forall t\in \mathbb{N}$ and let $e_1$ be the first canonical basis vector of $\mathbb{R}^d$, ie, $e_1 = (1,0,\cdots,0)$. Let us form a Gram Matrix $$\sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T $$

If I have a know a complete eigenspace decomposition of the above gram matrix, can I relate it to the unperturbed gram matrix $$\sum_{t=1}^Tx_tx_t^T $$ More specifically if I know $U,\Lambda$ s.t $$U\Lambda U^T = \sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T$$ and if $\tilde{U}, \tilde{\Lambda}$ are s.t. $$\tilde{U}\tilde{\Lambda}\tilde{U}^T = \sum_{t=1}^Tx_tx_t^T$$. Is there any relation that $U \approx \tilde{U}$ and $\Lambda \approx \tilde{\Lambda}$?

Let $x_t \in \mathbb{S}^{d-1}$, $\forall t\in \mathbb{N}$ and let $e_1$ be the first canonical basis vector of $\mathbb{R}^d$, ie, $e_1 = (1,0,\cdots,0)$. Let us form a Gram Matrix $$\sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T $$

If I have a know a complete eigenspace decomposition of the above gram matrix, can I relate it to the unperturbed gram matrix $$\sum_{t=1}^Tx_tx_t^T $$ More specifically if I know $U,\Lambda$ s.t $$U\Lambda U^T = \sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T$$ and if $\tilde{U}, \tilde{\Lambda}$ are s.t. $$\tilde{U}\tilde{\Lambda}\tilde{U}^T = \sum_{t=1}^Tx_tx_t^T$$. Is there any relation that $U \approx \tilde{U}$ and $\Lambda \approx \tilde{\Lambda}$?

EDIT: Okay, for example let us suppose $U=\tilde{U}$ that is the eigen vectors are same. And let us suppose $e_1$ is perpendicular to all but $u_1$ eigenvector corresponding to the maximum eigenvalue. Then $$\big(\sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T\big)u_i = (\sum_{t=1}^Tx_tx_t^T)u_i + (\sum_{t=1}^Te_1e_1^T)u_i + (\sum_{t=1}^Te_1x_t^T)u_i + (\sum_{t=1}^Tx_te_1^T)u_i$$ $$\implies \lambda_i(V_T) = \lambda_i(\tilde{V}_T) + 0 + 0$$ for all $i\neq 1$ Is this correct?

Let $x_t \in \mathbb{S}^{d-1}$, $\forall t\in \mathbb{N}$ and let $e_1$ be the first cannonical basis vector of $\mathbb{R}^d$, ie, $e_1 = (1,0,\cdots,0)$. Let us form a Gram Matrix $$\sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T $$

If I have a know a complete eigenspace decomposition of the above gram matrix, can I relate it to the unperturbed gram matrix $$\sum_{t=1}^Tx_tx_t^T $$ More specifically if I know $U,\Lambda$ s.t $$U\Lambda U^T = \sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T$$ and if $\tilde{U}, \tilde{\Lambda}$ are s.t. $$\tilde{U}\tilde{\Lambda}\tilde{U}^T = \sum_{t=1}^Tx_tx_t^T$$. Is there any relation that $U \approx \tilde{U}$ and $\Lambda \approx \tilde{\Lambda}$?

Let $x_t \in \mathbb{S}^{d-1}$, $\forall t\in \mathbb{N}$ and let $e_1$ be the first canonical basis vector of $\mathbb{R}^d$, ie, $e_1 = (1,0,\cdots,0)$. Let us form a Gram Matrix $$\sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T $$

If I have a know a complete eigenspace decomposition of the above gram matrix, can I relate it to the unperturbed gram matrix $$\sum_{t=1}^Tx_tx_t^T $$ More specifically if I know $U,\Lambda$ s.t $$U\Lambda U^T = \sum_{t=1}^T(x_t + e_1)(x_t+e_1)^T$$ and if $\tilde{U}, \tilde{\Lambda}$ are s.t. $$\tilde{U}\tilde{\Lambda}\tilde{U}^T = \sum_{t=1}^Tx_tx_t^T$$. Is there any relation that $U \approx \tilde{U}$ and $\Lambda \approx \tilde{\Lambda}$?