For which values of t does a matrix not have eigenvalues

Question

I need help solving this problem "For which values of real parameter t does the matrix:

\begin{bmatrix} π^2t^2 & 36\\ -36 & 0 \\ \end{bmatrix}

NOT have real eigenvalues. Thank you.

Answer 1

If we compute the characteristic equation we find that the eigenvalues are the solutions to $$\begin{align} -\lambda(\pi^2t^2-\lambda)+36^2 &= \lambda^2-\lambda \pi^2t^2 +36^2\\ &= 0 \end{align}$$ Using the quadratic formula we find that $$\lambda=\frac{\pi^2t^2\pm\sqrt{\pi^4t^4-4\cdot 36^2}}{2}$$ The eigenvalues are real if and only if $$\pi^4t^4-4\cdot 36^2>0$$ $$t^4>\frac{4\cdot36^2}{\pi^4}$$ Square rooting both sides $$t^2<-\frac{2\cdot 36}{\pi^2}\quad\text{or}\quad t^2>\frac{2\cdot 36}{\pi^2}$$ Again square rooting $$|t|<i6\pi\sqrt{2}\quad\text{or}\quad |t|>6\pi\sqrt{2}$$ If you are dealing with real $t$ this means that $t\in(-\infty,-6\pi\sqrt{2})\cup (6\pi\sqrt{2},\infty)$.

Answer 2

Hint: calculate the determinant of \begin{bmatrix} π^2t^2 - \lambda & 36\\ -36 & -\lambda \\ \end{bmatrix}, the result will be a second order polynomial in $\lambda$, set it to zero, and try ABC formula (specificly, check when the determinant > 0).