Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Algebra / 1 / c

Expanding the determinant along the second column givesIf $b\notin \{-3,2\}$, there are three distinct real eigenvalues. Part (b)(ii) gives a nonzero eigenvector for each, and part (b)(i) makes them independent. They form a basis, so the matrix is diagonalisable.

When $b=-3$,has rank one for every $a$: the third row is nonzero, and the second row is $a/5$ times the third. Therefore the eigenspace for $-3$ has dimension two. The eigenvalue $2$ provides at least one further independent eigenvector by part (b), so again there is an eigenbasis.

When $b=2$,Its rank is one if $a=0$ and two otherwise. Thus the eigenspace for $2$ has dimension two exactly when $a=0$. For $a\ne 0$ it has dimension one, while the eigenspace for $-3$ also has dimension one, since $X+3I$ then has rank two. There are no other eigenvalues by the characteristic polynomial. A diagonal matrix in an eigenbasis would need three independent eigenvectors, whereas these two eigenspaces supply only two; diagonalisation is impossible. Conversely, at $a=0$ their dimensions add to three and part (b) makes their sum direct. ThereforeThe argument uses only the determinant, rank-nullity, the established direct-sum result, and the definition of diagonalisation.