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Let $\mathbb{R}$ denote the field of the real numbers. Let $U=(u_1,u_2,...,u_n)\in (\mathbb{R}^n)^n$ be a base for $\mathbb{R}^n$ (i.e. $u_1,u_2,...,u_n$ are a base for $\mathbb{R}^n$). Let $\pi_U^{(i)}:\mathbb{R^n}\rightarrow\mathbb{R}$ be the projection that returns the $u_i$-component (for example, $\pi_U^{(i)}(\sum_{j=1}^n\alpha_j\,u_j)=\alpha_i)$. Let $A\in Mat_n(\mathbb{R})$. Define $\phi_U(A)$ to be the element of $Mat_n(\mathbb{R})$ whose first row is: $$(\pi_U^{(1)}(Au_1),\pi_U^{(1)}(Au_2),...,\pi_U^{(1)}(Au_n))$$ Second row: $$(\pi_U^{(2)}(Au_1),\pi_U^{(2)}(Au_2),...,\pi_U^{(2)}(Au_n))$$ Third row: $$(\pi_U^{(3)}(Au_1),\pi_U^{(3)}(Au_2),...,\pi_U^{(3)}(Au_n))$$ $$.$$$$.$$$$.$$ n-th row:$$(\pi_U^{(n)}(Au_1),\pi_U^{(n)}(Au_2),...,\pi_U^{(n)}(Au_n))$$

It follows easily that $\phi_U$ is a ring isomorphism of $Mat_n(\mathbb{R})$.

Question 1: Is $\{\phi_U | U$ is a base for $\mathbb{R}^n\}$ a subgroup of $Aut_{{Ring}}(Mat_n(\mathbb{R}))$ ? ($Aut_{Ring}$ means the group of ring isomorphisms)

Question 2: Is $Aut_{{Ring}}(Mat_n(\mathbb{R}))=$$\{\phi_U | U$ is a base for $\mathbb{R}^n\}$

Thank you

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  • $\begingroup$ @EwanDelanoy My linear algebra background is weak. I am now self-studying linear algebra, thus I cant answer your question. Question: Does $A\rightarrow (P^{-1} AP)^{T}$ preserve multiplication ? Remember that $\phi_U$ is a ring isomorphism $\endgroup$ – Amr Nov 16 '13 at 15:49
  • $\begingroup$ This should be helpful: math.stackexchange.com/questions/188684/… $\endgroup$ – Rasmus Nov 16 '13 at 15:50
  • $\begingroup$ @EwanDelanoy It's OK take your time $\endgroup$ – Amr Nov 16 '13 at 15:55
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Given any two bases $A=(a_1,\ldots,a_n)$ and $B=(b_1,\ldots,a_n)$ of ${\mathbb R}^n$, we can express the vectors of $A$ in terms of the vectors in $B$ : we have $a_j=\sum_{i=1}^n c_{ij}b_i$ where the $c_{ij}$ are real numbers. One then defines the matrix

$$ [A]_B=(c_{ij})_{1\leq i,j\leq n} $$

It is easy to check, using the definition of matrix multiplication, that for any three bases $A,B,C$, we have

$$ [A]_{C}=[B]_{C}[A]_{B} \tag{1} $$

With those notations, your $\phi_U(A)$ is simply $[AU]_{U}$. Taking $C=A$ in (1), we see that

$$ [B]_A=([A]_B)^{-1} \tag{2} $$

Denote by $E=(e_1,e_2,\ldots ,e_n)$ the canonical basis of ${\mathbb R}^n$. For convenience, let us still denote by $U$ the matrix $[U]_E$. Then

$$ \phi_U(A)=[AU]_{U}=[E]_U[AU]_{E}=[E]_UAU=U^{-1}AU \tag{3} $$

It follows easily from (3) that $\phi(UV)=\phi(U)\phi(V)$, so $\phi$ is a group homomorphism from $(GL_n({\mathbb R}),\times)$ to ${\sf Aut}({\sf Mat}_n({\mathbb R}))$.

The answer to question 1 is therefore YES (your set is the image of the homomorphism $\phi$, so it is certainly a subgroup).

Your question $2$ asks if this homomorphism $\phi$ is surjective. The answer to that one is YES also, and there are several ways to see this.

The question provided in Rasmus’ comment gives elegant, advanced answers.

If you prefer an elementary (and therefore longer and more involved) solution, here is a sketch of the proof :

Let $\phi$ be an automorphism of $({\sf Mat}_n({\mathbb R}))$. We want to show that $\phi(A)=U^{-1}AU$ for some $U$ as in (3). Denote by $(E_{ij})$ the canonical basis of ${\sf Mat}_n({\mathbb R})$, satisfying for any triple $1 \leq i,j,k \leq n$,

$$ E_{ij}e_{k}=\delta_{jk}e_i \tag{4} $$

and the fundamental property for any $1 \leq i,j,k,l \leq n$

$$ E_{ij}E_{kl}=\delta_{jk}E_{il} \tag{5} $$

where $\delta_{jk}$ is the Kronecker symbol. Let us put $F_{ij}=\phi(E_{ij})$. We deduce from (5) that for any $1 \leq i,j,k,l \leq n$

$$ F_{ij}F_{kl}=\delta_{jk}F_{il} \tag{6} $$

and then with a little work, (6) implies that there is a basis $F=(f_1,f_2,\ldots, f_n)$ of ${\mathbb R}^n$ such that for any $1 \leq i,j,k \leq n$ ,

$$ F_{ij}f_{k}=\delta_{jk}f_i \tag{7} $$

Then we may take $U=[F]_{E}$.

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  • $\begingroup$ Hi Ewan. Thanks for the time/effort that you put in writing this long answer. I will read it sometime during the next 2-3 days (sorry for that) and reply with my feedback. Thank you a lot, I appreciate your effort $\endgroup$ – Amr Nov 16 '13 at 18:27

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