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

A linear endomorphism is diagonalizable over its field precisely when its minimal polynomial is a product of distinct linear polynomials over that same field. Splitting alone is insufficient: repeated factors permit nontrivial Jordan blocks.

For the displayed matrix, a determinant expansion givesThese formulas can be reduced in each requested characteristic. The four cases below distinguish failure to split from failure to have distinct factors.

Over the complex numbers, the three eigenvalues areThey are distinct. Each is a root of the minimal polynomial, so here the minimal polynomial equals the characteristic polynomial and has distinct linear polynomials. Consequently $A$ is diagonalizable over $\mathbb{C}$.

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 $A$ is not diagonalizable over $\mathbb{R}$.

In the finite field ${\mathbb{F}}_3$, reduce the characteristic polynomial to $(x-1{)}^3$. For $B=A-I$, reduction of the entries givesHenceIts repeated linear polynomial prevents diagonalizability. Equivalently, a diagonalizable map with only the eigenvalue one would have to be $I$, and this matrix is not $I$.

In the finite field ${\mathbb{F}}_2$, the matrix becomesOn the first two coordinate vectors it is a nonzero nilpotent Jordan block of size two, and on the last it is the identity map. ConsequentlyThe repeated zero root prevents diagonalizability. Directly, the only possible eigenvalues are zero and one, so diagonalizability would imply $A^2=A$, whereas the displayed off-diagonal entry disappears on squaring.