What is the closed form expression of the largest eigenvalue of the matrix $\Lambda A\Lambda$?

Question

I came across the following problem

What is the closed form expression of the largest eigenvalue of matrix $P=\Lambda A\Lambda$ where \begin{align} \Lambda & = \text{diag} ( \lambda_1,...., \lambda_n),\\ A & = I_n - \sigma uu^T, \end{align} for some unit vector $u\in \mathbb R^n$ ($||u||=1$) and parameters $\sigma \in (0,1)$ and $\lambda_i>0\forall i$.

I have tried a lot of ideas, but doesn't work. See my attempts below. Could anyone give some hint? Thank you so much!

My attempts.

1. The first idea I have is to observe that if $\{v_1=u, v_2,..., v_n\}$ forms an orthonormal basis, then $A=V D V^T$ where $D=\text{diag} ( 1-\sigma, 1,..., 1)$ and $V = [v_1,...., v_n]\in \mathbb R^{n\times n}$. Now, we have $P= \Lambda V D (\Lambda V)^T$. Unfortunately, $\Lambda V \neq V\Lambda$, thus this expression of $P$ is not an eigendecomposition. Thus, we couldn't find the largest eigenvalue using this expression of $P$.

2. The second idea I found comes from the fact that the largest eigenvalue of matrix $P$ is the maximal value of the following optimization problem $\max_{x: ||x||\leq 1} \langle x, Px\rangle$. But I have no idea how to solve this optimization problem. An equivalent geometry result is that the largest eigenvalue of $P$ is exactly the half of generalized diameter of the ellipsoid $\{x \in \mathbb R^n: \langle x, P^{-1} x\rangle\leq 1\}$. But again, I have no intuition about this ellipsoid.

3. So I tried to google a found a more general question here. Unfortunately, the answer says that it is only possible for matrices of size $2\times 2$.