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

Choose a basis and let $A$ represent $T$. Define the characteristic polynomial byUnder a change of basis with invertible change matrix $S$, the new matrix is $S^{-1}AS$. HenceThus the polynomial is independent of the chosen coordinates. It is monic of degree $\dim V$.

To prove the sum is direct, suppose ${\sum }_{j=1}^r{v}_j=0$ with $v_j\in V_{{\lambda }_j}$. Fix $i$ and apply the polynomial in the linear operatorEvery $v_j$ with $j\ne i$ is killed by one factor. On $v_i$, each factor is multiplication by ${\lambda }_i-{\lambda }_j$, givingThe scalar coefficient is nonzero because the eigenvalues are distinct. Thus $v_i=0$ for every $i$, provingIn particular, choosing a basis in each eigenspace yields a linearly independent combined list.

If ${\chi }_T(\lambda )=0$, then $\det (\lambda I-A)=0$, so $\lambda I-A$ is singular. By the rank-nullity theorem it has a nonzero kernel vector $v$. ThusThis is exactly the definition of an eigenvalue. Here $\lambda$ lies in the scalar field: for the stated real vector space it is real. A nonreal root instead gives an eigenvalue of the complexified operator, by the same argument over $\mathbb{C}$.

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.