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Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$ and $S(\varepsilon)$ (which are always orthogonal and diagonal as matrices respectively) such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) U(\varepsilon)^T$ and the functions $U$ and $S$ are differentiable at $0$.

A stronger statement is likely possible. Also, we are not assuming that the eigenvalues are pairwise distinct.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

[edit] The non-negative assumption I had earlier was unnecessary. Forget. Further correction: The analysis version was over-complicated at the start.

Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$ and $S(\varepsilon)$ (which are always orthogonal and diagonal as matrices respectively) such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) U(\varepsilon)^T$ and the functions $U$ and $S$ are differentiable at $0$.

A stronger statement is likely possible. Also, we are not assuming that the eigenvalues are pairwise distinct.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$ and $S(\varepsilon)$ (which are always orthogonal and diagonal as matrices respectively) such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) U(\varepsilon)^T$ and the functions $U$ and $S$ are differentiable at $0$.

A stronger statement is likely possible. Also, we are not assuming that the eigenvalues are pairwise distinct.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

[edit] The non-negative assumption I had earlier was unnecessary. Forget.

Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$ and $S(\varepsilon)$ (which are always orthogonal and diagonal as matrices respectively) such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) U(\varepsilon)^T$ and the functions $U$ and $S$ are differentiable at $0$.

A stronger statement is likely possible. Also, we are not assuming that the eigenvalues are pairwise distinct.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

[edit] The non-negative assumption I had earlier was unnecessary. Forget. Further correction: The analysis version was over-complicated at the start.

Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$, $V(\varepsilon)$, $S(\varepsilon)$ such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) V(\varepsilon)^T$ is always an SVD and the functions $U$, $S$ and $V$ are differentiable at $0$.

A stronger statement is likely possible.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

[edit] The non-negative assumption I had earlier was unnecessary. Forget.

Let the dual numbers be $\mathbb R[\varepsilon]/(\varepsilon^2)$. Write a general dual number as $a + b \varepsilon$ (where $\varepsilon^2 = 0$). Given a symmetric matrix $M$ over the dual numbers (i.e. where $M = M^T$), it is known that $M$ can be written as $M = USU^T$ where $U^T = U^{-1}$ and $S$ is a unique diagonal dual-number matrix. $S$ is unique up to permutation of the diagonal entries. [1]

It is known that the dual numbers have a relationship to analysis (especially differentiation). This is especially clear if you use big-oh notation. You then have that $(a + b \varepsilon)(c + d\varepsilon) = ac + \varepsilon (ad + bc) + O(\varepsilon^2)$, which is essentially dual number multiplication. Therefore there is an analysis version of the above theorem that can be expressed using O notation.

Is there an analysis-type proof of the above spectral theorem? I imagine the statement is then:

Given a smooth symmetric-matrix-valued function $M(\varepsilon)$, there exist matrix-valued functions $U(\varepsilon)$ and $S(\varepsilon)$ (which are always orthogonal and diagonal as matrices respectively) such that $M(\varepsilon) = U(\varepsilon) S(\varepsilon) U(\varepsilon)^T$ and the functions $U$ and $S$ are differentiable at $0$.

A stronger statement is likely possible. Also, we are not assuming that the eigenvalues are pairwise distinct.

[1] - Generalisations of Singular Value Decomposition to dual-numbered matrices. This is published in a journal, but the Arxiv version is better.

[edit] The non-negative assumption I had earlier was unnecessary. Forget.