Eigenvalues of a submatrix A self-adjoint matrix $A$ in a complex linear space has real eigenvalues bounded between $0$ and $1$.
If $A$ is projected onto an arbitrary two-dimensional subspace, what would be the bounds on the corresponding two eignvalues? I am almost sure they will also be between $0$ and $1$ as well, but need to prove it.
EDIT: Here is a more explicit formulation of what is meant by projection:
I take two arbitrary orthonormal column vectors $x$ and $y$ , and make the following $2 \times 2$ matrix:
$$ B =\begin{pmatrix} x^{\dagger} A x & x^{\dagger} A y  \\   y^{\dagger} A  x &  y^{\dagger} A  y   \end{pmatrix}$$ 
Orthonormal means $x^{\dagger} x =  y^{\dagger} y =1$ and $x^{\dagger} y =0$.
What are the bounds on the eigvevalues of $B$?

Let $a = x^\dagger Ax$, $b = y^\dagger Ay$, and $c = x^\dagger Ay$. The eigenvalues of $B$ are determined by 
$$\lambda_\pm =\frac{a+b\pm\sqrt{(a-b)^2+4\vert c\vert ^2}}{2}.$$ 
Since $0\leq a,b\leq 1$ by assumption, then having $0\leq \lambda_{\pm}\leq1$ is equivalent to 
$$(a+b)^2\geq (a-b)^2+4\vert c\vert^2\qquad (1)$$ 
and
$$ \left(2-a-b\right)^2\geq \left(a-b\right)^2+4\vert c\vert^2.\qquad (2)$$
It is straightforward to show (1) using Cauchy-Schwarz with vectors $A^{1/2}x$ and $A^{1/2}y$. For (2), after simplifying the equivalent expressions we obtain is 
$$(1-a)(1-b)\geq \vert c\vert^2.\qquad(2')$$ 
Note that $1-a = x^\dagger\left(I-A\right)x$, $1-b = y^\dagger\left(I-A\right)y$, and $c = -x^\dagger\left(I-A\right)y$. Therefore, (2') is also provable by a simple application of Cauchy-Schwarz with vectors $\left(I-A\right)^{1/2}x$ and $\left(I-A\right)^{1/2}y$.
