The determinant of $T: \mathbb C^2\to \mathbb C^2$ as an $\mathbb R$-linear operator

Question

Suppose the determinant of a $\mathbb C$-linear transformation $T:\mathbb C^2\to \mathbb C^2$ is $a+bi$. I'm trying to prove that when $\mathbb C^2$ is identified with $\mathbb R^4$, the determinant of the $\mathbb R$-linear transformation $T:\mathbb R^4\to \mathbb R^4$ is $a^2+b^2$.

I started off with a lower dimensional case: a complex-linear operator $T': \mathbb C\to \mathbb C$ must be multiplication by a complex number $a+bi$, and if I write the matrix of $T'$ w.r.t. the basis $(1,i)$ of $\mathbb C$ over $\mathbb R$, then the determinant of the matrix is $a^2+b^2$.

In higher dimensional case, the complex-linear $T$ is multiplication by a $2\times 2$ matrix $$A=\begin{bmatrix}x_1+ix_2&z_1+iz_2\\y_1+iy_2&w_1+iw_2\end{bmatrix}.$$ The set $((1,0)^T,(i,0)^T,(0,1)^T,(0,i)^T)$ would be an $\mathbb R$-basis of $\mathbb C^2$. W.r.t. this basis, the matrix of $T$ is $$A'=\begin{bmatrix}x_1&-x_2&z_1&-z_2\\x_2&x_1&z_2&z_1\\y_1&-y_2&w_1&-w_2\\y_2&y_1&w_2&w_1\end{bmatrix}.$$

Am I supposed to compute the determinant of this last matrix and make sure it equals $a^2+b^2$ provided $\det A=a+bi$? It seems like a lot of calculations.

Answer 1

Let do this in full generality. Consider a matrix $Z=A+iB$ of shape $n\times n$ with entries in $\Bbb C$, isolate then $A,B$ from it with entries in $\Bbb R$. Then $Z$ induces a linear map $\Bbb C^n\to\Bbb C^n$, thus it induces by using the forgetful functor "$!$" a map, $$ !Z:(\Bbb R^2)^n\to (\Bbb R^2)^n\ , $$ and thus a matrix, after fixing some identification $\Bbb R^{2n}\to\Bbb R^{2n}$, and we are considering the determinant of this map.

Using the basis generalized from the one in case $n=2$ in the posted question, we get the matrix $$ !Z= \begin{bmatrix} A & -B\\B&A \end{bmatrix} \ , $$ and we want to compute its determinant. (With respect to some/any basis). Let $E$ (German notation, it is here better than the international notation $I$,) be the unit matrix of shape $n\times n$. We have the following equalities between block matrices: $$ \begin{aligned} \underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2} \underbrace{\begin{bmatrix} E \\ -iE\end{bmatrix}}_{2\times 1} &= \underbrace{\begin{bmatrix} E \\ -iE\end{bmatrix}}_{2\times 1} \underbrace{\begin{bmatrix} A +iB \end{bmatrix}}_{1\times 1} \ , \\ \underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2} \underbrace{\begin{bmatrix} E \\ iE\end{bmatrix}}_{2\times 1} &= \underbrace{\begin{bmatrix} E \\ iE\end{bmatrix}}_{2\times 1} \underbrace{\begin{bmatrix} A -iB \end{bmatrix}}_{1\times 1} \ , \qquad\text{(conjugated version)}\\ & \qquad\text{so putting all together in one block matrix computation}\\ \underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2} \underbrace{\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}}_{2\times 2} &= \underbrace{\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}}_{2\times 2} \underbrace{\begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix}}_{2\times 2} \ , \\ &\qquad\text{so we have after base change with $\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}$ the similitude} \\ \begin{bmatrix} A & -B\\B&A \end{bmatrix} &\sim \begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix} \ ,\qquad\text{so} \\ \det\begin{bmatrix} A & -B\\B&A \end{bmatrix} &= \det\begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix} \\ &=\det(A+iB)\cdot\det(A-iB) \\ &=\det(A+iB)\cdot\overline{\det(A+iB)} \\ &=|\ \det(A+iB)\ |^2 \ . \end{aligned} $$ Note: Structurally this means the following. We start with $\require{AMScd}$ \begin{CD} !\Bbb C^n @>!Z>> !\Bbb C^n\\ @V \cong V V @VV \cong V\\ \Bbb R^{2n} @>> W> \Bbb R^{2n} \end{CD} where $W$ is a matrix that we write down when the choice of the base change matrix for the two same vertical isomorphisms is fixed. Here, $\det W$ does not depend on the two same $\cong$ arrows.

We need $\det W$. The idea is to extend once more the field of scalars, from $\Bbb R$ to $\Bbb C$! The determinant remains. (This is similar to the fact, that the determinant of a matrix with entries in $\Bbb Q$ is the same one, if we consider all entries to be in $\Bbb R$ or $\Bbb C$...)

So we formally tensor over $\Bbb R$ with $\Bbb C$. The functorially induced diagram is $\require{AMScd}$ \begin{CD} @. !\Bbb C^n\otimes_{\Bbb R}\Bbb C @>!Z\otimes\operatorname{id}>> !\Bbb C^n\otimes_{\Bbb R}\Bbb C\\ @. @V \cong V V @VV \cong V\\ \Bbb C^{2n} @= \Bbb R^{2n}\otimes_{\Bbb R}\Bbb C @>> W> \Bbb R^{2n}\otimes_{\Bbb R}\Bbb C @= \Bbb C^{2n} \end{CD} But now we are free to choose the matrix $W$ using the $\cong$ arrows with entries in $\Bbb C$. And it turns out that we can make the choice such that over $\Bbb C$: $\require{AMScd}$ \begin{CD} !\Bbb C^n\otimes_{\Bbb R}\Bbb C @>!Z\otimes\operatorname{id}>> !\Bbb C^n\otimes_{\Bbb R}\Bbb C\\ @V \cong V V @VV \cong V\\ \Bbb C^{2n} @>> \begin{bmatrix} Z&\\&\bar Z\end{bmatrix}> \Bbb C^{2n} \end{CD} (Note: Such simple linear algebra "computations" may become structurally important, e.g. when studying Hodge structures... This is the only reason for the categorial overkill, that would be misplaced without this connection.)

Answer 2

Here is a terrible and lazy proof. If $W$ is a complex vector space, write $W^!$ for $W$ considered as a real vector space.

Given an element $\lambda$ of $\mathbb{C}^\times$, pick any $T$ with that element as its determinant (clearly such a $T$ exists), and consider the determinant of the linear transformation $T$ acting on the underlying real vector space $V^!$. I claim that this new determinant does not depend on the choice of $T$, but only on the element $\lambda$. Indeed, the map induced by $T$ on $\bigwedge^4 V^!$ is the same as the map induced by $\lambda$ on $\bigwedge^2 (\bigwedge^2 V)^!$ under the canonical isomorphism between the latter and $\bigwedge^4 V^!$.

It follows that we may check the claim on transformations of the form $$ \begin{pmatrix} \lambda & 0 \\ 0 & 1\\ \end{pmatrix} $$ for which it reduces to the case you already checked.

Answer 3

The map responsible for sending $A \mapsto A'$ is actually a group isomorphism $\boldsymbol{\Psi}: \operatorname{GL}(n, \mathbb{C}) \to \operatorname{GL}(2n, \mathbb{R})$. Moreover, $\det$ is a group homomorphism from $\operatorname{GL}(2n, \mathbb{R}) \to \mathbb{R}$.

The elementary matrices are the generators of the general linear group. Let $E \in \operatorname{GL}(n, \mathbb{C})$ be any elementary matrix.

• If $E$ is a column-switched matrix, it is easy to show $|\det(E)| = \det(\boldsymbol{\Psi}(E)) = 1$.

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \mapsto \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

The matrix on the left has two columns swapped, so the determinant is $-1$. The one on the right has two pairs of columns swapped, so the determinant is $1$.

• If $E$ is a column-added matrix, it is easy to show $|\det(E)| = \det(\boldsymbol{\Psi}(E)) = 1$.

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & x+yi & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \mapsto \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & x & -y & 1 & 0 & 0 & 0 \\ 0 & 0 & y & x & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

The matrix on the right trivially has determinant $1$. Now here is the meat of the problem.

• If $E$ is a column-multiplied matrix with factor $x+yi$, it is easy to show $|\det(E)| = \det(\boldsymbol{\Psi}(E)) = x^2+y^2$.

$$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x+yi & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \mapsto \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x & -y & 0 & 0 \\ 0 & 0 & 0 & 0 & y & x & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

Doing the normal determinant expansion, it becomes clear the determinant on the matrix on the right is $x\cdot x - (-y)\cdot y = x^2+y^2$.

Thus, $\det \circ \boldsymbol{\Psi} = |\det|$ on the generators of $\operatorname{GL}(n, \mathbb{C})$, which extends to the entire space.

Thus, if $M \in \operatorname{GL}(n, \mathbb{C})$ satsifies $\det(M) = a+bi$, then

$$\det(\boldsymbol{\Psi} M) = |\det M| = a^2 + b^2.$$

Answer 4

Assume w.l.o.g. that $A$ is in Jordan normal form, specifically, that it is upper triangular: $$A = \begin{bmatrix} \lambda_1 & * \\ 0 & \lambda_2 \end{bmatrix}$$ so that we have $\det A = \lambda_1\lambda_2 = a+ib$. Its eigenvalues can, of course, be complex: $\lambda_j=x_j+iy_j$, $x,y\in\mathbb R$. The corresponding real matrix is then block upper-triangular: $$A' = \begin{bmatrix} C_1 & * \\ 0 & C_2 \end{bmatrix}$$ with $$C_j = \left[\begin{array}{lr} x_j & -y_j \\ y_j & x_j \end{array}\right], \det C_j = \lambda_j\overline\lambda_j.$$ We then have $$\det A' = \det C_1 \det C_2 = \lambda_1 \overline\lambda_1 \lambda_2 \overline\lambda_2 = (\det A) (\overline{\det A}) = (a+ib)(a-ib) = a^2+b^2.$$

Answer 5

Suppose $A \in \mathbb{C}^{n \times n}$. Define $\phi:\mathbb{R}^{2n} \to \mathbb{C}^n $ as $\phi((x_1,y_1,...,x_n,y_n)) = \sum_k (x_k+iy_k) e_k$. It is straightforward to see that $\phi$ is linear and invertible (the inverse being real linear).

Write the basis $e_1,...,e_{2n} $ of $\mathbb{R}^{2n}$ as $u_1,v_1,...,u_n,v_n$.

The corresponding real basis of $\mathbb{C}^n$ is $e_1 = \phi(u_1),ie_1= \phi(v_1),...,e_n=\phi(u_n),ie_n=\phi(v_n)$.

We want to compute $\det \tilde{A}$, where $\tilde{A} = \phi^{-1}\circ A \circ \phi$.

If $V$ is invertible, note (some expansion needed) that $\det(\phi^{-1}\circ V^{-1}A V \circ \phi) = \det \tilde{A}$, hence we can choose $A$ to have whatever form suits, in this case the Jordan normal form.

Note that \begin{eqnarray} \tilde{A}(\alpha u_k+\beta v_k) &=& \phi^{-1}(A (\alpha u_k+i\beta v_k)) \\ &=& \phi^{-1}(\alpha \operatorname{re}(A e_k) - \beta \operatorname{im}(A e_k)+ i [ \alpha \operatorname{im}(A e_k) + \beta \operatorname{re}(A e_k) ] ) \\ &=& \phi^{-1}( \sum_i (\alpha \operatorname{re}[A]_{ik} - \beta \operatorname{im}[A]_{ik}) e_i + (\alpha \operatorname{im}[A]_{ik} + \beta \operatorname{re}[A]_{ik} ) i e_i) \\ &=& \sum_i (\alpha \operatorname{re}[A]_{ik} - \beta \operatorname{im}[A]_{ik}) u_i + (\alpha \operatorname{im}[A]_{ik} + \beta \operatorname{re}[A]_{ik} ) v_i \end{eqnarray} In particular, the representation of the map from the subspace spanned by $u_k,v_k$ to the coordinates of $u_i,v_i$ is given by the block matrix ${\bf \tilde A}_{ik} = \begin{bmatrix} \operatorname{re}[A]_{ik} & -\operatorname{im}[A]_{ik} \\ \operatorname{im}[A]_{ik} & \operatorname{re}[A]_{ik} \end{bmatrix}$ Hence, if $A$ is in Jordan normal form, then $\tilde{A}$ is an upper triangular block matrix, with each block being a $2 \times 2$ real matrix of the above form. Furthermore, the diagonal blocks are of the form $\begin{bmatrix} \operatorname{re} \lambda_k & -\operatorname{im} \lambda_k \\ \operatorname{im} \lambda_k & \operatorname{re} \lambda_k \end{bmatrix}$, where $\lambda_k$ are the eigenvalues of $A$.

Hence $\det {\tilde A} = \prod_k \det \begin{bmatrix} \operatorname{re} \lambda_k & -\operatorname{im} \lambda_k \\ \operatorname{im} \lambda_k & \operatorname{re} \lambda_k \end{bmatrix}= \prod_k |\lambda_k|^2$.

Since $\det A = \prod_k \lambda_k$, we have the desired result.