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I have been reading about calculating derivatives of eigenvalues for non-simple cases and it is not as easy as I hoped. But, then I realized that I don't need to calculate the derivatives for what I wanted to prove. So, I came up with the following proposition and its proof. But, after all I have read I'm not sure that I'm not violating any continuity or some other conditions in the proof. So the question is am I missing something important in the proof?

Proposition. Let $A(t) \in M_n(\mathbb{C})$ be analytical, Hermitian and positive definite for all $t \in [0, \infty)$. Let $A(0) = I_n$ and $\dot{A}(0)$ be negative definite, where $\dot{A}(t) = dA(t)/dt$. Then, there exists a $\bar{t} \in \mathbb{R}^{>0} \cup \{\infty\}$ such that $\rho(A(t)) < 1$ for all $t \in (0, \bar{t})$ where $\rho(\cdot)$ is the spectral radius.

Proof. Note that $\rho(A(0)) = 1$. Because of the continuity of eigenvalues, it is sufficient to show that $\dot{\lambda}(0) < 0$ (?) where $$A(t) x(t) = \lambda(t) x(t) ~~\text{and}~~ x^H(t) x(t) = 1.$$ Taking the derivative and multiplying by $x^H(t)$ from left, we obtain $$\begin{align} \dot{\lambda}(t) x(t) + \lambda(t) \dot{x}(t) &= \dot{A}(t) x(t) + A(t) \dot{x}(t) \\ \dot{\lambda}(t) x^H(t) x(t) + \lambda(t) x^H(t) \dot{x}(t) &= x^H(t) \dot{A}(t) x(t) + x^H(t) A(t) \dot{x}(t) \\ \dot{\lambda}(t) + \lambda(t) x^H(t) \dot{x}(t) &= x^H(t) \dot{A}(t) x(t) + \lambda(t) x^H(t) \dot{x}(t) \\ \dot{\lambda}(t) &= x^H(t) \dot{A}(t) x(t). \end{align}$$ Since $\dot{A}(0) < 0$ it follows that $\dot{\lambda}(0) < 0$ regardless of the eigenvectors selected.

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    How did you take the derivative? – 2017-01-02
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    @polfosol See my edited question. – 2017-01-02

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Your proof is incomplete. You have assumed, without proof, the differentiability of eigenvalues as well as the eigenvectors (Kato's classic, Perturbation Theory for Linear Operators, should be useful in this regard; this MO post is useful too).

There is actually no need to assume that. Recall that spectral norms of Hermitian matrices are equal to spectral radii. Since $A(t)$ is differentiable at $t$, by the definition of $\dot{A}(t)$, we have $\|A(t)-I-t\dot{A}(0)\|=o(t)$. Therefore \begin{align*} \rho(A(t))=\|A(t)\|&\le\|I+t\dot{A}(0)\|+o(t)\\ &=1+\lambda_\max\left(\dot{A}(0)\right)t+o(t)\tag{1}\\ &<1 \end{align*} when $t$ is sufficiently small (in particular, equality $(1)$ holds because all eigenvalues of $I+t\dot{A}(0)$ are positive when $t$ is small).

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    I'm a little confused right now. In the MO answer it says "If the symmetric matrix depends analytically upon one parameter, then you can follow analytically its eigenvalues and its eigenvectors. Notice that this requires sometimes that the eigenvalues cross. When this happens, the largest eigenvalues, as the maximum of smooth functions, is only Lipschitz.", but the given example in the question obviously contradicts this. – 2017-01-06
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    Also, it is known that eigenvalues and eigenvectors are differentiable when the eigenvalues are simple for all $t$. What if I can add the condition that the eigenvalues of $A(t)$ are simple when $t>0$? Does this make a difference? – 2017-01-06
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    I'm unfamiliar with the derivative definition you gave. How can we obtain that from $$\dot{A}(0) := \lim_{t \to 0} \frac{A(t) - I}{t}$$ Also it looks like you assumed $\lVert A(t) \rVert \geq \lVert I + t \dot{A}(0) \rVert$ in the inequality as the original should be $|\lVert A(t) \rVert - \lVert I + t \dot{A}(0) \rVert| \leq o(t)$. – 2017-01-06
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    @obareey (1) I don't see any contradiction in the MO post. $\lambda_1=1-t$ and $\lambda_2=1+t$ are analytic in $t$, but $\lambda_\max=\max(\lambda_1,\lambda_2)=1+|t|$ is not differentiable at $0$. (2) When crossing occurs, the largest eigenvalue is *in general* only Lipschitz, but it can be differentiable or analytic in specific cases (e.g. for constant function). (3) Let $H(t)=A(t)-I-t\dot{A}(0)$. Then $\|H(t)\|=o(t)$ and $\|A(t)\|=\|I+t\dot{A}(0)+H(t)\|\le\|I+t\dot{A}(0)\|+\|H(t)\|$. This is just triangle inequality. – 2017-01-06
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    (4) What your proof lacks is a justification of the differentiability of the eigenvalues and eigenvectors. If you assume that all eigenvalues of $A$ are simple in some interval $[0,T]$, then yes, your proof works, but as my answer shows, we don't actually need $\lambda(t),x(t)$ or $\rho(A(t))$ to be differentiable. – 2017-01-06