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Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2 = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2(t) = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2 = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2 = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Let $A$ be $C^\infty$. Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2 = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n)$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably).

(I chose the $2\times2$ example $A$ to be pointwise invertible for $t$ close to $0$, because this was asked for in the original question. But the pointwise invertibility of $A$ is not relevant: every differentiability problem that can occur for the eigenvalue $0$ can also occur for other eigenvalues.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.

Let us consider functions $A$ from (an open interval in) $\mathbb{R}$ into the set of symmetric real $n\times n$ matrices (Hermitian complex $n\times n$ matrices behave analogously).

If $A$ is given by $A(t) = diag(1+t,1-t)$, then the eigenvalue functions $\lambda_1,\lambda_2$ of $A$ with $\lambda_1\leq\lambda_2$ are $\lambda_1(t) = 1-|t|$ and $\lambda_2 = 1+|t|$, hence are not differentiable. Instead of differentiability of the ordered tuple of eigenvalues, we should therefore discuss the question whether there is a differentiable function $(\lambda_1,\dots,\lambda_n):\mathbb{R}\to\mathbb{R}^n$ that consists pointwise of the eigenvalues of $A$ counted with multiplicities (i.e.: can the eigenvalue functions be chosen differentiably?).

(I chose the $2\times2$ example $A$ to be pointwise positive definite for $t$ close to $0$, because this was asked for in the original question. But this is not relevant: every differentiability problem that can occur for any eigenvalue $\leq0$ can also occur for positive eigenvalues. Moreover, considering $A^tA$ instead of $A$ does not change any differentiability issue: If, for some pointwise positive definite $A$, the eigenvalues of $A^tA = A^2$ are not [resp. cannot be chosen] as regular as $A^tA$ is, then they are not [resp. cannot be chosen] as regular as $A$, because $A^tA$ and $A=\sqrt{A^tA}$ have the same regularity, due to the real-analyticity of $B\mapsto\sqrt{B}$.)

Some of the results of Alekseevsky/Kriegl/Losik/Michor: Choosing roots of polynomials smoothly and Kriegl/Michor: Differentiable perturbation of unbounded operators, or older results cited therein, are the following:

• If $A$ is $C^1$, then the eigenvalue functions $\lambda_1,\dots,\lambda_n$ can be chosen $C^1$ (cf. Kato: Perturbation theory for linear operators, §II.6.3, Theorem 6.8).
• If $A$ is real-analytic, then the eigenvalue functions (and also eigenvector functions) can be chosen real-analytically.
• If $A$ is $C^\infty$, then the eigenvalue functions can be chosen twice differentiably.
• Even if $A$ is $C^\infty$, the eigenvalue functions cannot always be chosen $C^2$ (example 7.4 in AKLM, first example in KM).
• Consider the eigenvalue functions $\lambda_1,\dots,\lambda_n$ with $\lambda_1\leq\dots\leq\lambda_n$ (they are always continuous). Assume for all $i,j\in\{1,\dots,n\}$ that either $\lambda_i=\lambda_j$ or there is no $t\in\mathbb{R}$ at which the functions $\lambda_i,\lambda_j$ meet of infinite order. Then the eigenvalue functions (and also eigenvector functions) can be chosen $C^\infty$.