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I am trying to solve the following problem: Let $$A^2=\begin{bmatrix} -2 & 2 & -4 \\ 2& 1 & -2\\ 4 &-6 & 6 \end{bmatrix}$$ Consider the trace of the matrix $A$ is $-1$. Then what is the determinant of the matrix $A$?

I have tried to solve the problem as follows:

Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues of the matrix $A$.

Then determinant is $\lambda_1\lambda_2\lambda_3$, but $\lambda_1^2\lambda_2^2\lambda_3^2=36 \implies \lambda_1\lambda_2\lambda_3=\pm 6.$

After that I am stuck.

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    $\begingroup$ You also know that $\lambda_1+\lambda_2+\lambda_3=-1$ since the sum of the eigenvalues equals the trace. $\endgroup$ – John Wayland Bales Jun 9 '16 at 17:24
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    $\begingroup$ The sum of their squares is equal to trace($A^2$) which is 5. $\endgroup$ – Aravind Jun 9 '16 at 17:33
  • $\begingroup$ Yeah but this does not help a lot. $\endgroup$ – user346772 Jun 9 '16 at 17:46
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Set $A^2 = B$. By writing and analyzing the characteristic polynomial of $B$, you can see that it has only one real root $\mu_1 > 0$ and hence two complex non-real conjugate roots $\mu_2, \mu_3 = \overline{\mu}_2$. Assuming $A$ is a real matrix with (possibly complex) eigenvalues $\lambda_i$, we can arrange the eigenvalues so that $\lambda_i^2 = \mu_i$. Since $\mu_2, \mu_3$ are not real, $\lambda_2, \lambda_3$ will also be not real and since $A$ is real, we will have $\lambda_1 \in \mathbb{R}$ and $\lambda_3 = \overline{\lambda}_2$. Now,

$$ \det(A) = \lambda_1 \lambda_2 \lambda_3 = \lambda_1 \lambda_2 \overline{\lambda}_2 = \lambda_1 |\lambda_2|^2 = \pm 6 $$

and so we will know the determinant if we will know the sign of $\lambda_1$.

Beyond this point, I'm not sure how we can continue without using the formula for the roots of a cubic or solving numerically. If we solve numerically, we see that $$\mu_1 \approx 6.91, \,\, \mu_2 \approx -0.95 + 2.07i, \,\, \mu_3 \approx -0.95 - 2.07i$$ (up to reordering $\mu_2,\mu_3$) which means we have two options for $(\lambda_2, \lambda_3)$ (up to reordering):

$$ \lambda_2 \approx 0.81 + 1.27i, \,\, \lambda_3 \approx 0.81 - 1.27i, \\ \lambda_2 \approx -0.81 - 1.27i, \,\, \lambda_3 \approx -0.81 + 1.27i $$

and $\lambda_1 \approx \pm 2.62$. In any case, the relation

$$ -1 = \lambda_1 + \lambda_2 + \lambda_3 = \lambda_1 + 2 \operatorname{Re}(\lambda_2) \approx \lambda_1 \pm 1.62 $$

implies that $\lambda_1$ must be negative and so $\det(A) = -6$.

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Finding the eigenvalues of $A^2$ numerically gives the real eigenvalue as around 6.9 and one complex eigenvalue around 0.95 + 1.25i. Taking square roots, the absolute values of the real eigenvalue of $A$ is around 2.6 and the real part of the complex eigenvalues has absolute value around 0.8. Since the trace is -1, this means the real eigenvalue of $A$ is negative; hence $det(A)=-6$.

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