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Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_nk$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example given by Denis Serre at Conjugation between commutative subalgebras of a matrix algebra?).

My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?

Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$.

Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_n(k)$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example given by Denis Serre at Conjugation between commutative subalgebras of a matrix algebra?).

My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?

Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$.

Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_nk$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example here).

My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?

Here by "close" I mean that for some small $\varepsilon>0$, we have $\|f(a) - g(a)\| \leq \varepsilon \|a\|$, for some submultiplicative norm on $A$.

Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_nk$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example given by Denis Serre at Conjugation between commutative subalgebras of a matrix algebra?).

My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?

Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$.

Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_nk$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example here).

My question is: Is this result still true if $f$ and $g$ are "close"? Here by "close" I mean that for some small $\varepsilon>0$, we have $\|f(a) - g(a)\| \leq \varepsilon \|a\|$, for some submultiplicative norm on $A$.

Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are conjugate in $M_nk$, i.e., there exists a matrix $X \in M_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.

It we do not assume that $A$ is simple, then this result is generally false (see e.g. the example here).

My question is: Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?

Here by "close" I mean that for some small $\varepsilon>0$, we have $\|f(a) - g(a)\| \leq \varepsilon \|a\|$, for some submultiplicative norm on $A$.