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Wikipedia states the following:

$\det \begin{pmatrix} A & B \\ B & A \end{pmatrix} = \det{(A+B)}\det{(A-B)}$

for $A$,$B$, square matrices of the same size.

Now I have a strong suspicion that the following also holds:

$\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix} = \det{(A+iB)}\det{(A-iB)}$

This obviously holds if $A,B$ are just numbers, and I've tested it in a few cases for $A,B$ of size $2\times 2$.

But how do you prove it? Usually one would try decomposing the matrix into two pieces which are upper or lower triangular, but I coulnd't come up with a suitable decomposition.

So I have two questions:

  1. How do you prove the first formula?
  2. Can you provide a proof (or counterexample) to the second formula?
glS
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John Donne
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1 Answers1

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Both formulaes are true. I will prove second formula only.

First, let us suppose that $A$ is invertible. Since

$$1\cdot\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det \begin{pmatrix} I & 0 \\ -CA^{-1} & I \end{pmatrix}\cdot\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det\begin{pmatrix} A & B \\ 0 & D-CA^{-1}B \end{pmatrix}$$ then $$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix} =\det(A)\cdot\det(D-CA^{-1}B)$$

Now $$\det\begin{pmatrix} A & -B \\ B & A \end{pmatrix}=^{1)}\det\begin{pmatrix} A+iB & iA-B \\ B & A \end{pmatrix}=^{2)}\det(A+iB)\det(A-B(A+iB)^{-1}(iA-B))=^{3)}$$ $$=^{3)}\det(A+iB)\det(A-iB(iA-B)^{-1}(iA-B))=\det(A+iB)\det(A-iB)$$

1) Add to row $1$ row $n+1$ multiplied by $i$, to row $2$ row $n+2$ multiplied by $i$, etc up to row $n$

2) use formula derived in the first part

3) $(kA)^{-1}=\frac1kA^{-1}$ for $k=i$

We have proved the formula for case when $A$ is invertible. Now we should get rid from this assumption. For any $t \in \mathbb{R}$ let $A_t=A-tI$. If $A_t$ is not invertible, then $t$ is an eigenvalue of $A$. Contrapositive give us the following statement: If $t$ is not an eigenvalue of $A$, then $A_t$ is invertible(and thus the formula is true). Let us fix arbitrary $A$ and $B$ and consider $$F(t)=\det\begin{pmatrix} A_t & -B \\ B & A_t \end{pmatrix}-\det(A_t+iB)\det(A_t-iB)$$ $F(t)$ is a polynomial in $t$. $F(t)=0$ for all $t$ except the finite amount of eigenvalues of $A$ - so $F(t)$ vanishes in infinitely many points. But nonzero polynomial can have only finite number of roots. So F(t) is zero polynomial, and $F(t)=0$ for any $t$, and we have to take $t=0$ to establish the formula.

First formula derivation should be very easy now, so I will skip it.

Hedgehog
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