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Let A be a $3\times 3$ invertible matrix (with real coefficients) and let $B=A^TA^{-1}$. Prove that

\begin{equation*} \det(I + B) = 2(1 + tr(B)). \end{equation*}

I know that

\begin{equation*} \det (I+B)=\lambda_1\lambda_2\lambda_3+\lambda_1\lambda_2 +\lambda_1\lambda_3+\lambda_2\lambda_3+\lambda_1+\lambda_2+\lambda_3+1 \end{equation*}

given $\lambda_1,\lambda_2$ and $\lambda_3$ are three distinct eigenvalues of $B$. However, I don't know where to go on from here and how to utilise the fact that $B=A^TA^{-1}$. Any help or direction would be appreciated.

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    $\begingroup$ Is it guaranteed that $B$ has three distinct eigenvalues? If it does, then it is diagonalizable and then working in a basis of eigenvectors for $B$, the problem becomes trivial. $\endgroup$ – Alex M. May 20 '15 at 16:01
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    $\begingroup$ Note that $\det B = 1$ $\endgroup$ – Ben Grossmann May 20 '15 at 16:06
  • $\begingroup$ It's not guaranteed. I was only going in that direction because I know that since B is 3x3, it cannot have more than 3 distinct eigenvalues. Should I throw the det(I + B) equation away then? $\endgroup$ – linearalgebrathrowaway May 20 '15 at 16:06
  • $\begingroup$ @linearalgebrathrowaway No, it's true regardless of whether or not the eigenvalues are distinct. $\endgroup$ – jgon May 20 '15 at 16:07
  • $\begingroup$ The statement in question can be viewed as a generalisation of the identity $\det(I+Q)\equiv 2(1+\operatorname{tr}(Q))$ for $Q\in SO(3;\mathbb R)$. In fact, when the symmetric part of $A$ is positive definite, $B$ is similar to some $Q$ in the special orthogonal group. $\endgroup$ – user1551 May 20 '15 at 16:55
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Note that $\det B = 1$.

Note additionally that $B^{-1} = A(A^T)^{-1}$, so that $$ \operatorname{trace}(B^{-1}) = \operatorname{trace}(A(A^T)^{-1}) = \operatorname{trace}((A^T)^{-1}A) = \operatorname{trace}(B^T) = \operatorname{trace}(B) $$ Now, your polynomial can be written as $$ \det(I + B) = \det(B) + \det(B)\operatorname{trace}(B^{-1}) + \operatorname{trace}(B) + 1 $$ the conclusion follows.

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  • $\begingroup$ I follow the majority of that; one quick question though, how do you know that det(B)trace(B^−1) = λ1λ2 + λ1λ3 + λ2λ3? $\endgroup$ – linearalgebrathrowaway May 20 '15 at 16:27
  • $\begingroup$ Write $$ \det(B) = \lambda_1 \lambda_2 \lambda_3\\ \operatorname{trace}(B^{-1}) = \lambda_1^{-1} + \lambda_2^{-1} + \lambda_3^{-1} $$ is it clear now? $\endgroup$ – Ben Grossmann May 20 '15 at 16:28
  • $\begingroup$ yes, thank you very much, appreciate it! $\endgroup$ – linearalgebrathrowaway May 20 '15 at 16:33
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Hint: $tr(A^{-1}A^T)=tr(A^TA^{-1})=tr(A^{-T}A)=tr(B^{-1})$

$\sum \lambda_i = tr(B)$

$\sum \lambda_1\lambda_2 = det(B)tr(B^{-1})$

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An alternative approach, which doesn't use the explicit formula in terms of eigenvalues. Instead, find the characteristic polynomial. See that, for any $\lambda$,

$$p(\lambda) = \det (B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \, B - \lambda x + \det B$$

for some number $x$. Here we merely have $\lambda = -1$.

The number $x$ can be computed in a variety of ways. Using exterior algebra is one method: construct the $3 \times 3$ matrix $B_2$, which acts on $3 \times 1$ column vectors corresponding to elements of $\bigwedge^2 \mathbb R^3$. Then $x = \mathrm{Tr} \, B_2$.

The relationship between $B_2$ and $B$ is explicitly

$$B_2 (a \wedge b) = B(a) \wedge B(b)$$

for any vectors $a, b$.

Now, use a common inversion identity:

$$B^{-1}(a) = \star B_2^T(\star a)/\det B$$

where $\star$ is the usual Hodge dual. This means we can write $B_2$ as

$$B_2(a \wedge b) = \star (B^T)^{-1}(\star [a \wedge b]) \det B$$

You can verify now (e.g. by breaking into a basis) that $\mathrm{Tr} \, B_2 = \det B \, \mathrm{Tr} \, (B^T)^{-1}$. By the arguments given in other answers, this is merely $\det B \, \mathrm{Tr} \, B$, and as a result, we have

$$p(\lambda) = \det(B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \, B - \lambda [\det B \, \mathrm{Tr} \, B] + \det B$$

For $\lambda =-1$, and since $\det B = 1$, the result follows.

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It suffices to show that $1+\lambda_1+\lambda_2+\lambda_3=\lambda_1\lambda_2\lambda_3+\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1$. But this is true because $\det(A^\top {A^{-1}})=1$, meaning $\lambda_1\lambda_2\lambda_3=1$.

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    $\begingroup$ You have not shown that $\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1=\lambda_1+\lambda_2+\lambda_3$. $\endgroup$ – jgon May 20 '15 at 16:12
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For my opinion, it seems more "comfortable" to use Newton's identities, especially when B is NOT invertible. e.g.$\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3=\dfrac{1}{2} [(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda^2_1+\lambda^2_2+\lambda^2_3)]=\dfrac{1}{2}(tr(B)^2-tr(B^2))$

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