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Let $A$ be a real $4 \times 4$ matrix such that $\operatorname{trace}(A) = 0$ and $A + xA^2 = I$, where $x \in \Bbb R$. Show that $$\det(A) = -\frac{16}{27}$$

I solved this by showing that all such matrices have a jordan form with that value for the determinant. But I expect there should be another less tedious method. Perhaps using Cayley-Hamilton theorem, but am unable to see how. Does anyone know how, or have seen a similar problem? (For $3 \times 3$ matrices, same constraints, $\det(A) = -1/4$.)

I came across this problem in Lovelock and Rund's book "Tensors, Differential Forms, and Variational Principles". It is problem 4.22 on page 129, but expressed using tensors. I think the authors expected a solution using generalised Kronecker deltas, but I got stuck, so used standard matrices. Also, can matrices of that form be diagonalised? How to show that?

Rodrigo de Azevedo
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Zamarath
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  • How did you solve it? – FShrike Jan 04 '22 at 12:26
  • Well, very tediously!! I let a, b, c, d be the main diagonal, and for the supra-diagonal in Jordan form I put 1,0,0 (from top left). Then I used the stated conditions, set up the required equations and solved them. There are eight cases, some contradict the given conditions, the others that are possible give det (A) = -16/27. Surely there is another way? – Zamarath Jan 04 '22 at 12:50
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    The two conditions imply that $x\ne0$ and the only possible eigenvalues of $A$ over $\mathbb C$ are the roots of the equation $x\lambda^2+\lambda-1=0$, namely $\lambda_\pm=\frac12\left(-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{4}{x}}\right)$. Use the given conditions to argue that the spectrum of $A$ must be ${\lambda_+,\lambda_+,\lambda_+,\lambda_-}$ or ${\lambda_-,\lambda_-,\lambda_-,\lambda_+}$. The trace condition thus becomes $3\lambda_\pm+\lambda_\mp=0$, i.e., $$ \frac{2}{x}=\pm\sqrt{\frac{1}{x^2}+\frac{4}{x}}. $$ – user1551 Jan 04 '22 at 16:46
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    ...Solve for $x$ and in turn obtain $x,\lambda_+$ and $\lambda_-$. Use the trace condition to find out which of ${\lambda_+,\lambda_+,\lambda_+,\lambda_-}$ or ${\lambda_-,\lambda_-,\lambda_-,\lambda_+}$ is the correct spectrum. Having determined the spectrum, you can compute the determinant. – user1551 Jan 04 '22 at 16:46
  • @user1551 Why is it the case that the given polynomial is the characteristic one? We surely only have that the given polynomial is a minimal polynomial, so all the roots are eigenvalues, but there could be more eigenvalues since the minimal polynomial is not the characteristic one. Characteristic polynomial of a $4$ dimensional matrix needs to be of order $4$, right? – FShrike Jan 04 '22 at 17:42
  • @FShrike The minimal polynomial must divide every annihilating polynomial. Therefore, the zeroes of the minimal polynomial is always a (possibly proper) subset of the zeroes of any given annihilating polynomial. – user1551 Jan 04 '22 at 17:59
  • @user1551 A subset indeed. I don’t know how you’re concluding that the only possible eigenvalues of $A$ are the roots of $x\lambda^2+\lambda-1$, when these roots can only be said to be a subset of the spectrum rather than the whole spectrum itself... – FShrike Jan 04 '22 at 18:02
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    My concern is that $x\lambda^2+\lambda-1$ is not the characteristic polynomial, so no conclusions can be made about all the eigenvalues. As in, you write $\Lambda={\lambda_+,\lambda_+,\lambda_-,\lambda_-}$ for example, but for all we know it could be: $\Lambda={\lambda_+,\lambda_-,a,b}$... @user1551 – FShrike Jan 04 '22 at 18:11
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    @FShrike Whether it is the characteristic polynomial (it isn't, because its degree is two) is unimportant. The key here is that it is an annihilating polynomial of $A$. – user1551 Jan 04 '22 at 18:14
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    @FShrike See the Wikipedia page for the minimal polynomial or this math.SE question; there is a stronger relationship than what you are claiming. – angryavian Jan 04 '22 at 19:10
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    @angryavian Thank you for that. It makes sense now that all the eigenvalues are roots of the minimal polynomial. – FShrike Jan 04 '22 at 19:35
  • @user1551 Thanks for your comment. I did not know about minimal polynomials! From your comment I can see that for any $n \times n$ matrix $A$ $(n > 2)$ with the given properties, and, say, $p$ positive entries on the main diagonal, it follows that $$ \det(A) = \frac{(2p-n)^{n} (-1)^{(n-p)}}{p^{p} (n-p)^{(n-p)}}$$ where $n-p \ne 0$ and $2p-n \ne 0$. For $n > 4$ the determinant values are not unique. – Zamarath Jan 06 '22 at 06:27
  • And the number of distinct values for $\det(A)$ is $\lfloor \frac{(n-1)}{2} \rfloor$. – Zamarath Jan 07 '22 at 09:20
  • A correction. In a previous comment, I should have said " $p$ positive entries on the main diagonal of a diagonal matrix similar to $A$". Also $A$ can be diagonalised since the minimal polynomial factors into distinct linear factors. – Zamarath Jan 13 '22 at 04:15

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