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Let $V = \mathbb{C}^2$. Let $T: V\to V$ denote a $\mathbb{C}$-linear transformation with determinant $a + bi$, $a,b\in \mathbb{R}$. Prove that if we consider $V$ as a 4-dimensional real vector space, then the determinant of $T$ as an $\mathbb{R}$-linear transformation is $a^2 + b^2$.

I don't think I am supposed to do this computationally. I would like to use a convenient formula for the determinant, e.g., as a product of eigenvalues, but I don't think I am allowed to in this case (I don't think it applies to all linear transformations $T$). I would appreciate any direction. Thanks,

LCrouch
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  • Hint: (a+bi)(a-bi) = a^2 + b^2. Also the characteristic polynomial of T over R4 hs real coefficients, so for every complex root c+di we have c -di is a foot as well. Work out one or two potential examples to see the structure of the matrix T over R4. – Kunal Chawla Jan 03 '20 at 01:58
  • This is a particular case of https://math.stackexchange.com/questions/1801911/determinant-of-determinant-is-determinant/1801957#1801957 , since the $4\times 4$-matrix that represents the $\mathbb{R}$-linear map $T$ is a block $2\times 2$-matrix with its four $2\times 2$-blocks mutually commuting. – darij grinberg Jan 03 '20 at 02:29

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I will prove this for arbitrary complex space (since $\dim V = 2$ is not really necessary here).

There is a matrix-based proof that goes in the following way. At first you take a $\mathbb{C}$-basis $E = \{e_1, \dots, e_n\}$ in $V$ and after that you have an $\mathbb{R}$-basis $E_r = \{e_1, \dots, e_n, ie_1, \dots, ie_n\} = E \bigcup iE$. Then as soon as operator $T:V \rightarrow V$ is $\mathbb{C}$-linear you have some structure in matrix $T(E_r)$ of your operator in basis $E_r$. It appears that $$ T(E_r) = \begin{bmatrix} Re T(E) & - Im T(E)\\\ Im T(E) & Re T(E) \end{bmatrix} $$ where $T(E)$ is the complex matrix of operator $T$ in original basis $E$. Now we apply some basic linear algebra: $$ 2^{2n}\det T(E_r) = \det \begin{bmatrix} I & iI\\\ I & -iI \end{bmatrix} \begin{bmatrix} Re T(E) & - Im T(E)\\\ Im T(E) & Re T(E) \end{bmatrix} \begin{bmatrix} I & I\\\ -iI & iI \end{bmatrix} $$ Here we used one simple fact: $$ \begin{bmatrix} I & iI\\\ I & -iI \end{bmatrix} \begin{bmatrix} I & I\\\ -iI & iI \end{bmatrix} = \begin{bmatrix} 2I & 0\\\ 0 & 2I \end{bmatrix} $$ And therefore by computing block multiplication we can get $$ 2^{2n}\det T(E_r) = \det \begin{bmatrix} 2T(E) & 0\\\ 0 & 2\overline{T(E)} \end{bmatrix} = 2^{2n} \det T(E) \det \overline{T(E)} = 2^{2n} |\det T(E)|^2 $$ So, we obtained equality$\det T(E_r) = |\det T(E)|^2$.

Matsmir
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