Let A= a b
c d
be a 2 x 2 matrix, where a,b,c and d are real numbers. We say that A admits an eigenvector if there exists a unit vector u and a real number s such that Au=su.
Show that A admits an eigenvector precisely when (tr A)^2 - 4detA >= 0
where detA is the determinant of A and trA is the trace of A ie the sum of the entries on the leading diagonal (the diagonal from the upper left to the lower right) of A.
On
For the $2\times 2$ matrix in question, let us denote the two eigenvalues as $\lambda_1$ and $\lambda_2$.
The characteristic polynomial of the matrix will be a quadratic in variable $s$, with roots equal to the eigenvalues $$(s-\lambda_1)(s-\lambda_2)=s^2-(\lambda_1+\lambda_2)s+\lambda_1\lambda_2$$
For the eigenvalues to be real, the discriminant of this polynomial must be greater than or equal to zero. The discriminant is given by $$(\lambda_1+\lambda_2)^2-4(1)(\lambda_1\lambda_2)$$
Now the trace of a matrix is the sum of its eigenvalues, while the determinant of a matrix is the product of the eigenvalues.
Denoting $Tr(A)$ as the trace of the matrix $A$, and $det(A)$ as the determinant, we have $$Tr(A)=\lambda_1+\lambda_2$$ $$det(A)=\lambda_1\lambda_2$$ Thus the discriminant will be $$Tr(A)^2-4det(A)$$ leading to the following condition to hold $$Tr(A)^2-4det(A)\geq 0$$
Note that $\det (A - \lambda) = \lambda^2 - (a+d)\lambda - bc$, when we solve for $\lambda$ using the quadratic formula, we get:
$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$ which has a real solution iff $\sqrt{(a+d)^2 - 4(ad-bc)} \geq 0$. This is equivalent to Trace$(A)^2 - 4 \det(A) \geq 0$.