Eigenvalue bound for quadratic maximization with linear constraint This builds on my earlier questions here and here.

Let $B$ be a symmetric positive definite matrix in $\mathbb{R}^{k\times k}$ and consider the problem
$$\begin{array}{ll} \text{maximize} & x^\top B x\\ \text{subject to} & \|x\|=1 \\ & b^\top x = a\end{array}$$
where $b$ is an arbitrary unit vector and $a > 0$ is a small positive number. Let $$\lambda_1 > \lambda_2 \geq \cdots \geq \lambda_k > 0$$ be the eigenvalues of $B$ with corresponding eigenvectors $z_1,...,z_k$. I conjecture that the optimal value of the problem is bounded below by $a^2 \lambda_1 + \left(1-a^2\right)\lambda_2$, at least if $a$ is small enough.

To motivate this conjecture, let us consider two special cases. First, suppose that $a= 0$. Then, as was explained to me in one of my previous posts, the optimal value is between $\lambda_1$ and $\lambda_2$ by the Courant-Fischer theorem. Thus, $\lambda_2$ is a lower bound, and it also coincides with my conjectured lower bound in this special case.
Second, let $a > 0$ but suppose that $b = z_i$ for some $i = 1,...,k$. Any feasible $x$ can be written as
$$x = ab + \sqrt{1-a^2} \cdot \hat{b}$$
where $\hat{b}\perp b$. If $b = z_1$, I can take $\hat{b} = z_2$, and if $b = z_i$ for $i \neq 1$, I can take $\hat{b} = z_1$. Either way, the objective value of $x$ is bounded below by $a^2 \lambda_1 + \left(1-a^2\right)\lambda_2$ as long as $a$ is small enough (note that this requires $\lambda_1 > \lambda_2$).
The difficulty is showing that it holds in the case where $b$ is not one of the eigenvectors of $B$ (perhaps with additional restrictions on how large $a$ can be). My intuition is that, if $b$ is not required to be orthogonal to $x$, but only "almost" orthogonal (meaning that $a$ may be required to be sufficiently small), you should be able to go a bit further in the direction of the principal eigenvector than in the case where $a = 0$.

Here is the most up-to-date work on this problem. In the answer below, it was found that the optimal value $v$ of the problem is a generalized eigenvalue of the system
$$PBx = vPx,$$
which in turn was derived from the system
$$PBPy + aPBb = v Py.$$
Any pair $\left(y,v\right)$ that solves these equations then leads to a feasible $x = ab+Py$, with $v$ being the objective value.
We can write
$$\left(vI - PB\right)Py = aPBb.$$
Note that, for any $v$ that is not an eigenvalue of $PB$, the matrix $vI-PB$ is invertible, whence
$$Py = a\left(vI-PB\right)^{-1}PBb.$$
The normalization $x^\top x = 1$ then becomes $y^\top P y = 1-a^2$, leading to the equation
$$\frac{1-a^2}{a^2} = b^\top BP\left(vI-PB\right)^{-2} PBb.$$
The largest root of this equation is the optimal value of the problem. Perhaps, as suggested, it can be found numerically.
 A: The following analysis explores various approaches to the problem, but ultimately fails to produce a satisfactory solution.
One of the constraints can be rewritten using the nullspace projector of $b$
$$\eqalign{
P &= \Big(I-(b^T)^+b^T\Big)
  = \left(I-\frac{bb^T}{b^Tb}\right)
  \;=\; I-\beta bb^T \\
Pb &= 0,\qquad P^2=P=P^T \\
}$$
and the introduction of an unconstrained vector $y$
$$\eqalign{
b^Tx &= a \\
x &= Py + (b^T)^+a \\
 &= Py + a\beta b \\
 &= Py + \alpha_0 b \\
}$$
The remaining constraint can be absorbed into the definition
of the objective function itself
$$\eqalign{
\lambda &= \frac{x^TBx}{x^Tx} 
 \;=\; \frac{y^TPBPy +2\alpha_0y^TPBb +\alpha_0^2\,b^TBb}{y^TPy +\alpha_0^2\,b^Tb}
 \;=\; \frac{\theta_1}{\theta_2} \tag{0} \\
}$$
The gradient can be calculated by a straightforward (if tedious)
application of the quotient rule as
$$\eqalign{
\frac{\partial\lambda}{\partial y}
 &= \frac{2\theta_2(PBPy +\alpha_0PBb)-2\theta_1Py}
   {\theta_2^2} \\
}$$
Setting the gradient to zero yields
$${
PBPy +\alpha_0PBb = \lambda Py \tag{1} \\
}$$
which can be rearranged into a generalized eigenvalue equation.
$$\eqalign{
PB\left(Py+\alpha_0b\right) &= \lambda Py \\
PBx &= \lambda Px  \tag{2} \\
}$$
Note that multiplying the standard eigenvalue equation
$$\eqalign{
Bx &= \lambda x  \tag{3} \\
}$$
by $P$ reproduces equation $({2})$. So both standard and generalized eigenvalues are potential solutions.
Unlike the discrete $\lambda$ values yielded by the eigenvalue methods,
equation $({1})$ is solvable for a continuous range of $\lambda$
$$\eqalign{
 y &= \alpha_0(\lambda P-PBP)^+PBb \\
}$$
and produces a $y$ vector which satisfies
the zero gradient condition $({1})$.
Unfortunately, none of these approaches yields a solution which satisfies all of the constraints.
But solving equation $(0)$ for an optimal $y$ vector is still the appropriate goal, and requires a numerical rather than an analytical approach.
A: I don't think the conjecture is correct. For a counter example, take $B=\begin{pmatrix} 1&0&0 \\0&1&0 \\ 0&0&\varepsilon \end{pmatrix}$ and $b=\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$. Then the desired maximum is $(1-a^2)+a^2\varepsilon < 1$.
