# Conjugation by the transpose homomorphism

Consider the map $\varphi: GL_n(K) \to M_n(K)$ from the set of all $n \times n$ invertible matrices to simply the set of all $n \times n$ matrices. Define this map by $\varphi(X) = AXA^T$, for some fixed matrix $A$. Is $\varphi$ a homomorphism if and only if $A^{-1} = A^T$?

I'm playing around with some Lie group ideas and the general idea for showing conjugation maps are indeed homomorphisms is to do something like $$AXYA^{-1} = AXA^{-1}AYA^{-1},$$ and then separate. To do this however, we would require that $A^{-1} = A^T$.

• What kind of homomorphism do you want? A group homomorphism? But then, what operation do you have in mind for $M_n(K)$ to make it a group? – verret Apr 6 '17 at 22:49

If $\phi$ is homomorphism (with operation on $M_n(K)$ being multiplication)
1) image is contained in $GL_n$: suppose some $X$ maps to noninvertible matrix, but $\phi(X)\phi(X^{-1})$ must be equal to $1$, therefore it's impossible
2) $A$ must be invertible, as $1 = 1^{-1} \Rightarrow \phi(1) = AA^T = (AA^T)^{-1}$, but $A$ is invertible iff $AA^T$ is
3) $\phi(1)$ should be equal $1$, so $\phi(1) = AA^T = 1 \Rightarrow A^T = A^{-1}$.
You may consider different situation: $$\sigma: GL_n(K) \times M_n(K) \rightarrow M_n(K), \quad (g, X) \to g^tXg$$ Then $\sigma$ is a homomorphism to a group $GL(M_n(K))$ which corresponds to action of $GL_n$ on bilinear forms on $K^n$.