The <quadratic polynomial> $x^2+4x+10=(x+2)^2+6$ has no real <roots of a polynomial>. Thus the <characteristic polynomial> does not split over the <real numbers>. A real <diagonal matrix> would have only real <eigenvalues>, and <similar matrices> have the same <characteristic polynomial>. Therefore \b[$A$ is not <diagonalizable> over $\mathbb R$].

Solution

Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 2 / a / ii / Solution

 Solution (source code)