Lower bound for lowest eigenvalue of sum of two special hermitian matrices. Assume you have a composition of hermitian matrices $H_{i}$ of the form,
$H=cos(\alpha)H_{1}+sin(\alpha)H_{2}$
with $\alpha\in[0,\tfrac{\pi}{2}]$ and you know the following things:
a) $H_{2}=V^{\dagger}H_{1}V$ is a unitary transformation of $H_{1}$.
b) The lowest eigenvector $v_{i}$ of $H_{i}$ has eigenvalue $\lambda_{i}$. Furthermore you the have the property that,
$$H_{i}v_{j}=\delta_{ij}\lambda_{j}v_{j}.$$
From this we can conclude,
$$v^{\dagger}_{i}v_{j}=\delta_{ij}.$$ Since both matrices related by a unitary mapping $\lambda_{1}=\lambda_{2}$.
c) The spectrum of $H_{i}$ is symmetric, meaning that the eigenvalues come in pairs $\{\lambda_{k},-\lambda_{k}\}$.
My intuition is somehow that the lowest eigenvalue $\lambda$ for the composed Matrix $H$ is bounded from below in the sense that it has to fullfill the property, that
$$\lambda\geq{\lambda_{1}},$$ where the equal sign only holds for $\alpha\in\{0,\tfrac{\pi}{2}\}$. 
Maybe one can weaken the condition a),c) in the sense that the matrices $H_{i}$ are not linked by a unitary transformation and the spectrum is not symmetric.
Actually im not able to prove this statement, also i don't have found an counter example, so i would also be happy about that =).
 A: The conjecture is not true. I wrote a simple python script which generates random matrices with the properties you want, and after a few runs it gave me a counter-example. Note that we may presume $H_1$ to be diagonal. $H_1$ is then defined by a random list of real numbers (which come in pairs), with smallest eigenvalue $\lambda_1$ and at least one zero eigenvalue, which I call $\lambda_2$. I then generate $H_2$ by swapping the eigenvectors with eigenvalues $\lambda_1$ and $\lambda_2$, and doing a random unitary on the rest of the spectrum. After a few runs, I then got the following result for the lowest eigenvalue as a function of $\alpha$:

I am adding the Python script, since it's a good idea to check I haven't made a mistake. Moreover, this script might be useful to play around with to possibly find some version of the statement that you might like to be true.
import numpy as np
from scipy.linalg import qr
import pylab as plt

chi = 20  #dimension of matrices, must be at least three

#generate diagonal Hermitian matrix H1 = D with the necessary properties
#(lambda_1 = lowest, lambda_2 = 0, lambda_n > lambda_1 for n > 1)
if chi%2==0:
    D = [np.random.random((1))[0] for i in range(chi/2-1)] + [0]
    D = D + [-d for d in D]
else:
    D = [np.random.random((1))[0] for i in range(chi/2)]
    D = D + [0] + [-d for d in D]
D = np.sort(np.array(D))
D = np.append(np.append(D[:1],D[chi/2:chi/2+1]),np.append(D[1:chi/2],D[chi/2+1:]))
H1 = np.diag(D)

#generate H2 by swapping the eigenvectors for lambda1 and lambda2,
#and do a random unitary on rest of spectrum
Ublock = np.random.random((chi-2,chi-2)) + 1j*np.random.random((chi-2,chi-2))
Ublock,_ = qr(Ublock)
X = np.array([[0,1],[1,0]])
U = np.zeros((chi,chi),dtype=complex)
U[0:2,0:2] = X
U[2:,2:] = Ublock
H2 = np.tensordot(np.tensordot(U,H1,(1,0)),U.conj().T,(1,0))

#check lowest eigenvalue as a function of alpha
alpha_list = np.linspace(0,np.pi/2,100)
lowest = []
for alpha in alpha_list:
    H = np.cos(alpha)*H1 + np.sin(alpha)*H2
    e = np.linalg.eigvals(H)
    e = np.sort(np.real(e))
    lowest.append(e[0])
plt.plot(alpha_list/np.pi,lowest)
plt.ylabel(r'lowest eigenvalue')
plt.xlabel(r'$\alpha/\pi$')
plt.savefig('plot.pdf')

