# Determinant of a linear map given by conjugation

Suppose we have two fields $K\subset L$ and $A\in GL(n,L)$. See $L^{n\times n}$ as a $K$-vectorspace, then $$C_A\colon L^{n\times n}\rightarrow L^{n\times n},B\mapsto ABA^{-1}$$ is a $K$-linear map. What is the determinant of $C_A$?

It seems trivial, but I'm too stupid right now. I could go into deep calculating to get a representation matrix for this linear map and compute the determinant of this matrix, but I'm sure that must work much easier.

Edit: Even if I just consider the example $n=2$, $K=\mathbb{R}$ and $L=\mathbb{C}$, I'm not getting any idea of the determinant without starting to calculate the representation matrix. Does someone have an idea at least to this specific example?

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Why does it seem trivial? –  Jonas Meyer Jan 11 '13 at 17:42
Just a feeling of mine. Conjugating with a fixed matrix seems like a very natural linear map... Do you think the best way is starting to calculate? –  max Jan 11 '13 at 17:47
Do you know how to get the matrix of the linear transformation (with respect to whatever bases you are interested in)? –  JohnD Jan 11 '13 at 18:25
For sure. Calculating the linear transformation is just some basic linear algebra, but I don't see how I can get enough information about this matrix with just abstract arguments to determine the determinant. –  max Jan 11 '13 at 18:27

$\def\GL{\operatorname{GL}}$ Lemma Let $K$ be a field and $A\in\GL(n,K)$. The map $\gamma_A^K:C\in M_n(K)\mapsto ACA^{-1}\in M_n(K)$, which is a $K$-linear map, has determinant $\delta^K(A)=1$.

Proof. Since $\gamma_A^K\circ\gamma_B^K=\gamma_{AB}^K$, we have $\delta^K(AB)=\delta^K(A)\delta^K(B)$, and then $\delta^K(A)=\delta^K(DAD^{-1})$ for all $D$. In other words, the function $\delta^K$ is invariant under conjugation in $\GL(n,K)$.

Now, if $A$ is diagonal with diagonal elements $\lambda_1$, $\dots$, $\lambda_n$, then the map $\gamma_A$ is $(c_{i,j})\mapsto(\lambda_i\lambda_j^{-1}c_{i,j})$. We see at once at the eigenvectors are the elementary matrices and that the eigenvalues are the numbers $\lambda_i\lambda_j^{-1}$. The determinant $\delta^K(A)$, which is the product of the eigenvalues, is in this case then easily seen to be $1$. We therefore have that $\delta^K(A)=1$ for all diagonal invertible matrices, and it follows from this that $\delta^K(A)=1$ for all diagonalizable invertible matrices because of conjugation invariance.

Now, if the field is algebraically closed, then the set of diagonalizable matrices is Zariski dense in the set of all invertible matrices, and the map $\delta$ is regular —it is a rational function in the coefficients— so $\delta$ is actually identically $1$.

If $K$ is not algebraically closed, let $\bar K$ be its algebraic closure. Then of course $A\in\GL(n,\bar K)$ and by what we have proved we have $\delta^{\bar K}(A)=1$. It is easy to see that $\delta^K(A)=\delta^{\bar K}(A)$, so we reach the conclusion we want also in this case. $\Box$

Lemma. Let $L/K$ be a finite field extension and let $N:L\to K$ be the norm, so that $N(l)$ is the determinant of the $K$-linear map $x\in L\mapsto lx\in L$. If $V$ is an $L$-vector space and $f:V\to V$ is an $L$-linear map, let $\det_Lf$ and $\det_Kf$ be the determinants of $f$ viewed as an $L$-linear map and as a $K$-linear map. Then $N(\det_Lf)=\det_Kf$.

Proof. We can suppose that $f$ preserves a complete flag $0=V_0\subset V_1\subset\cdots\subset V_n=V$ of $L$-subspaces of $V$; if not, we can extend scalars to a finite extension $L'/L$ which this does happen, and check that the end result is the same.

Then $\dim_LV_i/V_{i-1}=1$ for all $1\leq i\leq n$, and $f$ induces an $L$-linear map $f_i:V_i/V_{i-1}\to V_i/V_{i-1}$ which is given by multiplication by some scalar $l_i\in L$. We have $\det_L=l_1\dots l_n$.

Now the $V_i$'s are of course also $K$-subspaces of $V$, forming now a partial flag in $V$ viewed as a $K$-vector space, and $f$ preserves it. The determinant of $f$ viewed as a $K$-linear map is then the product of the determinants of the maps $f_i$ viewed themselves as $K$-linear maps. Since we know that $f_i$ is essentially multiplication by $l_i$ on $L$, by definition, $\det_Kf_i=N(l_i)$. The lemma follows from this. $\Box$

In this proof, we are computing the determinant of a matrix by first taking it to a block upper diagonal form, so to speak. We can now do what we really wanted:

Proposition. Let $K\subseteq L$ be a finite field extension and let $A\in\GL(n,L)$. Then the determinant of the $K$-linear map $C\in M_n(L)\mapsto ACA^{-1}\in M_n(L)$ is $1$.

Proof. By the second lemma, the determinat in question is the norm for the extension $L/K$ of the determinant of the same map viewed as an $L$-linear map. We computed this last determinant to be $1$ in the first lemma, and $N(1)=1$. $\Box$

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Over $\mathbb C$ we can replace the Zarisky topology by the usual topology, and the same reasoning works. –  Mariano Suárez-Alvarez Jan 11 '13 at 18:57
The morale of this is that for a lot of purposes we can pretend that all matrices are diagonal :-) –  Mariano Suárez-Alvarez Jan 11 '13 at 19:01
I don't follow. Suppose $n=2,\,K=\mathbb{R}$ and $L=\mathbb{C}$. As a vec. sp. over $\mathbb{R}$, $\mathbb{C}^{2\times2}\simeq\mathbb{R}^8$. So the $\mathbb{R}$-linear map $C\mapsto ACA^{-1}$ should have 8 eigenvalues. However, for your $\gamma_A$, there are only 4 eigenvalues. It seems to me that you are considering the $\mathbb{C}$-linear map $C\mapsto ACA^{-1}$ defined on $\mathbb{C}^{2\times2}$ rather than the natural extension of a $\mathbb{R}$-linear map $C\mapsto ACA^{-1}$ defined on $\mathbb{R}^8$ to a $\mathbb{C}$-linear map defined on $\mathbb{C}^8$. –  user1551 Jan 11 '13 at 20:47
If $A$ is a matrix with coefficients in a field $K$ and $L\supseteq K$ is an extension of $K$, consider the map $\gamma_A^L:C\in M_n(L)\mapsto ACA^{-1}\in M_n(L)$, which is an $L$-linear map as and such it has a determinant $\det\gamma_A^L$. The problem is to determine $\det\gamma_A^K$. What I am saying is that $\det\gamma_A^L=\det\gamma_A^K$ for all fields $L\supseteq K$, so we can take $L$ to be the algebraic closure of $K$. –  Mariano Suárez-Alvarez Jan 11 '13 at 21:01
(I have no idea why you think I was considering what you call the «natural extension», since it is quite unnatural to do so :-) ) –  Mariano Suárez-Alvarez Jan 11 '13 at 21:02

your question is so good hint:i think there are two solution 1)compute matrix of this linear transformation respect to standard base then obtain determinate of this matrix 2) or compute eigen value of this linear transformation and obtain determinate of this matrix

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Yeah, this is what I called "deep calculating". I am searching for a more elegant solution. Maybe there is no, but I don't think so... –  max Jan 11 '13 at 18:22