Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 2 / b

Expansion down the last column gives the characteristic polynomialThe quadratic polynomial has roots $(1\pm \sqrt{5})/2$, which are not rational numbers. For example, the rational root theorem restricts any rational root of this monic integer polynomial to $1$ or $-1$, and neither works. If $A$ had a diagonalization over $\mathbb{Q}$, its characteristic polynomial would be a product of linear factors with rational numbers as their roots. It is not, so $A$ is not diagonalizable over $\mathbb{Q}$.

Over the complex numbers, the eigenvaluesare distinct. Thus distinct eigenvalues imply diagonalizability, and $A$ is diagonalizable over $\mathbb{C}$, in fact already over $\mathbb{R}$.

An explicit eigenbasis isFor either quadratic root $\lambda$, the identity ${\lambda }^2=\lambda +1$ verifies $A{v}_{\lambda }=\lambda v_{\lambda }$. Taking these three columns as $P$ gives $P^{-1}AP=\mathrm{diag}(1,{\lambda }_{+},{\lambda }_{-})$. Their distinct eigenvalues ensure linear independence.

In the finite field ${\mathbb{F}}_5=\mathbb{Z}/5\mathbb{Z}$, the characteristic polynomial becomesbecause $(\lambda -3{)}^2={\lambda }^2-\lambda -1$ modulo $5$. Splitting into linear factors does not by itself guarantee diagonalizability: the repeated eigenvalue needs enough eigenvectors.

Solving $(A-3I)(x,y,z{)}^T=0$ givesso $y=3x$ and $z=4x$. Hencehas dimension one, although $3$ has algebraic multiplicity two. The geometric multiplicity is too small, and $A$ is not diagonalizable over ${\mathbb{F}}_5$. This illustrates diagonalizability depends on the field: characteristic zero gave three distinct eigenvalues, while reduction modulo $5$ merges two of them.