3 little mistake field; also k needs not be infinite; added 104 characters in body

I think it's a matter of basic linear algebra.

Generally, let $k$ be an infinite a field, and $L$ be some finite-dimensional $k$-algebra. (In our case, $k=\mathbb C$ and $L=\mathrm{M}_n\left(k\right)$.) If $A\in L$ is invertible, then $A^{-1}$ lies in the Zariski closure of the set $\left\lbrace 1,A,A^2,A^3,...\right\rbrace$.

Proof. Let $P\in k\left[L\right]$ be a polynomial such that $P\left(A^i\right)=0$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well. Here we naively consider $k\left[L\right]$ as denote the algebra of all polynomial functions from $L$ to $k$ (since where a "polynomial function" means a function that can be written as a polynomial in the coordinates). (If $k$ is infinite, this is , of course, isomorphic to the non-naive algebra of coordinate functions)functions, i. e. the symmetric algebra $\mathrm{S}\left(L^{\ast}\right)$, but we don't care about this isomorphy and therefore we don't need $k$ to be infinite.)

Let $P\in k\left[L\right]$ be a polynomial such that $P\left(A^i\right)=0$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well.

Define a $k$-algebra $U$ by $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ as a vector space, but with the multiplication being inherited from $L$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $L$, but as an algebra it is a direct product!

Let $N=\deg P$. Then the polynomial $P:L\to k$ can be written as $P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$, where $s:L\to U$ is the canonical map given by

$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$,

and $p:U\to k$ is some $k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our $U$, but we are only using the vector space structure on $U$, and as I said, as a vector space $U$ is just the tensor algebra of $L$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.N$.) Now consider the element$s\left(A\right)\in U$. This element$s\left(A\right)$is invertible (since$A$is invertible, so that$A^{\otimes i}$is invertible for every$i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N L^{\otimes i}$ is componentwise), and the algebra$U$is finite-dimensional (although its dimension is usually quite large). Thus,$s\left(A\right)^{-1}$lies in the$k$-linear span of the set$\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$(because if$u$is an invertible element of some finite-dimensional$k$-algebra, then$u^{-1}$lies in the$k$-linear span of the set$\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional$k$-algebra is algebraic over$k$). Since$s$is a$k$-algebra homomorphismmultiplicative map, we have$\left(s\left(A\right)\right)^i=s\left(A^i\right)$for all$i$, so that this becomes: The element$s\left(A^{-1}\right)$lies in the$k$-linear span of the set$\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since$p$is a linear map, we can apply$p$here and obtain: The element$p\left(s\left(A^{-1}\right)\right)$lies in the$k$-linear span of the set$\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now$p\circ s=P$, so this becomes: The element$P\left(A^{-1}\right)$lies in the$k$-linear span of the set$\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when$P\left(A^i\right)=0$for all$i\in\mathbb N$, then$P\left(A^{-1}\right)=0$, qed. 2 slight generalization I think it's a matter of basic linear algebra. Generally, if let$A\in\mathrm{GL}_n\left(k\right)$for some k$ be an infinite field, and $k$, L$be some finite-dimensional$k$-algebra. (In our case,$k=\mathbb C$and$L=\mathrm{M}_n\left(k\right)$.) If$A\in L$is invertible, then$A^{-1}$lies in the Zariski closure of the set$\left\lbrace 1,A,A^2,A^3,...\right\rbrace$. Proof. Let$P\in k\left[\mathrm{M}_n\left(k\right)\right]$k\left[L\right]$ be a polynomial such that $P\left(A^i\right)$ P\left(A^i\right)=0$for every$i\in\mathbb N$. We must then prove that$P\left(A^{-1}\right)=0$as well. Here we naively consider$k\left[\mathrm{M}_n\left(k\right)\right]$k\left[L\right]$ as the algebra of all polynomial functions from $\mathrm{M}_n\left(k\right)$ L$to$k$(since$k$is infinite, this is, of course, isomorphic to the non-naive algebra of coordinate functions). Define a$k$-algebra$U$by $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes L^{\otimes i}$ as a vector space, but with the multiplication being inherited from$\mathrm{M}_n\left(k\right)$L$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $\mathrm{M}_n\left(k\right)$, L$, but as an algebra it is a direct product! Let$N=\deg P$. Then the polynomial$P:\mathrm{M}_n\left(k\right)\to P:L\to k$can be written as$P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes L^{\otimes i}$, where$s:\mathrm{M}_n\left(k\right)\to s:L\to U$is the canonical map given by$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$, and$p:U\to k$is some$k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our$U$, but we are only using the vector space structure on$U$, and as I said, as a vector space$U$is just the tensor algebra of$\mathrm{M}_n\left(k\right)$L$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.

Now consider the element $s\left(A\right)\in U$. This element $s\left(A\right)$ is invertible (since $A$ is invertible, so that $A^{\otimes i}$ is invertible for every $i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes L^{\otimes i}$ is componentwise), and the algebra $U$ is finite-dimensional (although its dimension is usually quite large). Thus, $s\left(A\right)^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$ (because if $u$ is an invertible element of some finite-dimensional $k$-algebra, then $u^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional $k$-algebra is algebraic over $k$). Since $s$ is a $k$-algebra homomorphism, we have $\left(s\left(A\right)\right)^i=s\left(A^i\right)$ for all $i$, so that this becomes: The element $s\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since $p$ is a linear map, we can apply $p$ here and obtain: The element $p\left(s\left(A^{-1}\right)\right)$ lies in the $k$-linear span of the set $\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now $p\circ s=P$, so this becomes: The element $P\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when $P\left(A^i\right)=0$ for all $i\in\mathbb N$, then $P\left(A^{-1}\right)=0$, qed.

1

I think it's a matter of basic linear algebra.

Generally, if $A\in\mathrm{GL}_n\left(k\right)$ for some infinite field $k$, then $A^{-1}$ lies in the Zariski closure of the set $\left\lbrace 1,A,A^2,A^3,...\right\rbrace$.

Proof. Let $P\in k\left[\mathrm{M}_n\left(k\right)\right]$ be a polynomial such that $P\left(A^i\right)$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well. Here we naively consider $k\left[\mathrm{M}_n\left(k\right)\right]$ as the algebra of all polynomial functions from $\mathrm{M}_n\left(k\right)$ to $k$ (since $k$ is infinite, this is, of course, isomorphic to the non-naive algebra of coordinate functions).

Define a $k$-algebra $U$ by $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$ as a vector space, but with the multiplication being inherited from $\mathrm{M}_n\left(k\right)$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $\mathrm{M}_n\left(k\right)$, but as an algebra it is a direct product!

Let $N=\deg P$. Then the polynomial $P:\mathrm{M}_n\left(k\right)\to k$ can be written as $P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$, where $s:\mathrm{M}_n\left(k\right)\to U$ is the canonical map given by

$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$,

and $p:U\to k$ is some $k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our $U$, but we are only using the vector space structure on $U$, and as I said, as a vector space $U$ is just the tensor algebra of $\mathrm{M}_n\left(k\right)$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.

Now consider the element $s\left(A\right)\in U$. This element $s\left(A\right)$ is invertible (since $A$ is invertible, so that $A^{\otimes i}$ is invertible for every $i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$ is componentwise), and the algebra $U$ is finite-dimensional (although its dimension is usually quite large). Thus, $s\left(A\right)^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$ (because if $u$ is an invertible element of some finite-dimensional $k$-algebra, then $u^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional $k$-algebra is algebraic over $k$). Since $s$ is a $k$-algebra homomorphism, we have $\left(s\left(A\right)\right)^i=s\left(A^i\right)$ for all $i$, so that this becomes: The element $s\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since $p$ is a linear map, we can apply $p$ here and obtain: The element $p\left(s\left(A^{-1}\right)\right)$ lies in the $k$-linear span of the set $\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now $p\circ s=P$, so this becomes: The element $P\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when $P\left(A^i\right)=0$ for all $i\in\mathbb N$, then $P\left(A^{-1}\right)=0$, qed.