Return to Answer

$\DeclareMathOperator\ker{ker}$If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to $$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $\ker{f}=\ker{g}$.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\varepsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $\ker{f}=\ker{g}$ whenever $\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$ for all $a\in A$.

Proof. If $A$ is nilpotent, there is nothing to prove: every homomorphism $A\to M_n(k)$ is either 1-to-1 or identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $\ker{f}\neq \ker{g}$. Then, either $p_i\subseteq \ker{f}\backslash \ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq \ker{g}\backslash \ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$.

$\DeclareMathOperator\ker{ker}$Assume that $A$ is semisimple, so by Wedderburn's theorem, $A$ is isomorphic to $$M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $\ker{f}=\ker{g}$.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\varepsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $\ker{f}=\ker{g}$ whenever $\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$ for all $a\in A$.

Proof. Let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $\ker{f}\neq \ker{g}$. Then, either $p_i\subseteq \ker{f}\backslash \ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq \ker{g}\backslash \ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$.

Edit/Correction: The original claim was incorrect. Rather than deleting the answer, we modified it to get some benefit, a partial answer out of what was written. Please see Yves de Cornulier's reply for a counterexample for non-semisimple $A$.

$\DeclareMathOperator\ker{ker}$If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to $$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $\ker{f}$ and $\ker{g}$ are isomorphic.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\varepsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $\ker{f}=\ker{g}$ whenever $\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$ for all $a\in A$.

Proof. If $A$ is nilpotent, there is nothing to prove: every homomorphism $A\to M_n(k)$ is either 1-to-1 or identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $\ker{f}\neq \ker{g}$. Then, either $p_i\subseteq \ker{f}\backslash \ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq \ker{g}\backslash \ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$.

$\DeclareMathOperator\ker{ker}$If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to $$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $\ker{f}=\ker{g}$.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\varepsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $\ker{f}=\ker{g}$ whenever $\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$ for all $a\in A$.

Proof. If $A$ is nilpotent, there is nothing to prove: every homomorphism $A\to M_n(k)$ is either 1-to-1 or identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $\ker{f}\neq \ker{g}$. Then, either $p_i\subseteq \ker{f}\backslash \ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq \ker{g}\backslash \ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$.

If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to $$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $ker{f}$ and $ker{g}$ are isomorphic.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\epsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $ker{f}=ker{g}$ whenever $\|f(a)-g(a)\|\leq \epsilon\|a\|_A$ for all $a\in A$.

Proof. If $A$ is nilpotent, there is nothing to prove: every homomorphism $A\to M_n(k)$ is either 1-to-1 or identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\epsilon = \frac{1}{2}\min\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\}.$$

Now suppose $ker{f}\neq ker{g}$. Then, either $p_i\subseteq ker{f}\backslash ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq ker{g}\backslash ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\epsilon\|p_i\|_A$.

$\DeclareMathOperator\ker{ker}$If we do not assume that $A$ is simple, by Wedderburn's theorem, $A$ is isomorphic to $$J\oplus M_{n_1}(k)\oplus M_{n_2}(k)\oplus\dots\oplus M_{n_r}(k)$$ where $n_1\leq n_2\leq\dots\leq n_r$ positive integers and $J$ is the Jacobson radical of $A$, which is the maximal nilpotent ideal of $A$. Thus, two homomorphisms $f,g:A\to M_n(k)$ are equivalent if and only if $\ker{f}$ and $\ker{g}$ are isomorphic.

Claim: Given a norm $\|.\|_A$ on $A$, there exists $\varepsilon>0$ such that for any two homomorphisms $f,g:A\to M_n(k)$, $\ker{f}=\ker{g}$ whenever $\|f(a)-g(a)\|\leq \varepsilon\|a\|_A$ for all $a\in A$.

Proof. If $A$ is nilpotent, there is nothing to prove: every homomorphism $A\to M_n(k)$ is either 1-to-1 or identically $0$. Otherwise, let $p_i$ denote the identity of $M_{n_i}(k)$ so that $\{p_1,\dots,p_r\}$ is a set of pairwise orthogonal projections such that $p_1+\dots+p_r=1$. Let $$\varepsilon = \frac{1}{2}\min\Big\{\frac{1}{\|p_1\|_A},\dots,\frac{1}{\|p_r\|_A}\Big\}.$$

Now suppose $\ker{f}\neq \ker{g}$. Then, either $p_i\subseteq \ker{f}\backslash \ker{g}$ for some $i=1,\dots,r$, or $p_i\subseteq \ker{g}\backslash \ker{f}$ for some $i=1,\dots,r$. In either case, $\|f(p_i)-g(p_i)\|=1>\varepsilon\|p_i\|_A$.