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say I have a linear space $V$ over $\Bbb C$ and a linear transformation $T:V \to V$ such that $T=A+iB$ where $A,B \in \Bbb R^{n \times n}$

I proved already that $T_\Bbb R = \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$ (which mean $T$ reduced to $\Bbb R$ when you look at $V$ as a linear space over $\Bbb R$).

I need to prove that $\det(T_\Bbb R) = |\det(T)|^2$. can someone help please?

izikgo
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1 Answers1

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From the answer to your other question here, we know that $$ \det(T_\mathbb{R})=\det(A+iB)\det(A-iB). $$ Now $T=A+iB$ can be seen as a complex matrix with real part $A$ and imaginary part $B$. Denote $\overline{T}=A-iB$.

You just have to convince yourself that $\det \overline{T}=\overline{\det T}$. You can see it directly from the Leibniz formula.

Therefore $$ \det(T_\mathbb{R})=\det T\cdot \det \overline{T}=\det T\cdot \overline{\det T}=|\det T|^2. $$

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Julien
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  • thanks! sorry I divided my question into 2 different posts I just thought it would be easy once I find what $\det (T_\Bbb R)$ is... is there another way to show that $\det \overline{T}=\overline{\det T}$ without leibniz formula? cause we didn't learn that... – izikgo Apr 22 '13 at 17:17
  • @izikgo What properties/definition of det do you know? – Julien Apr 22 '13 at 18:58
  • just the one with the minors – izikgo Apr 22 '13 at 19:40
  • @izikgo But a minor is a determinant...So do you mean you have a definition by induction on the dimension? – Julien Apr 22 '13 at 19:42
  • yes. Laplace expansion. – izikgo Apr 22 '13 at 20:43
  • @izikgo So when you have to compute an $n\times n$ determinant, you get down to $n-1\times n-1$, then $n-2\times n-2$, until you reach $1\times 1$? I hope you will be presented with a more tractable approach soon...The Laplace expansion is convenient to compute some determinants, but it is not a very interesting "definition". – Julien Apr 22 '13 at 21:05