= Solution
Expanding the <determinant> along the second column gives
$$
\chi_X(t)=(t+3)(t-b)(t-2).
$$
If $b\notin\{-3,2\}$, 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 $b=-3$,
$$
X+3I=\begin{pmatrix}0&0&0\\2a&0&a\\10&0&5\end{pmatrix}
$$
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 $b=2$,
$$
X-2I=\begin{pmatrix}-5&0&0\\2a&0&a\\10&0&0\end{pmatrix}.
$$
Its rank is one if $a=0$ and two otherwise. Thus the <eigenspace> for $2$ has dimension two exactly when $a=0$. For $a\neq0$ 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. Therefore
$$
\boxed{X_{a,b}\text{ is diagonalisable}\quad\Longleftrightarrow\quad b\neq2\text{ or }a=0.}
$$
The argument uses only the <determinant>, rank-nullity, the established direct-sum result, and the definition of diagonalisation.
Back to article page
