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

The primary decomposition theorem for linear maps states that ifwith distinct monic irreducible polynomials $p_j$, thenEach summand is a $T$-invariant subspace. More generally the same direct sum conclusion holds for any factorization of an annihilating polynomial into pairwise coprime polynomials, using the corresponding kernels. The underlying mechanism is the Bézout's identity for polynomials, which separates the mutually coprime factors.

Since $T$ is diagonalizable, its minimal polynomial is ${\prod }_{j=1}^r(x-{\lambda }_j)$ with distinct eigenvalues ${\lambda }_j$ in the field. Apply the primary decomposition theorem for linear maps:For a scalar that is not an eigenvalue, the associated eigenspace is zero, so the sum indexed by the whole field has only finitely many nonzero summands. The eigenspaces need not be one-dimensional.

For $v\in V_{\lambda }$, use commutation to obtainThus $Sv\in V_{\lambda }$, proving that every $T$-eigenspace is an $S$-invariant subspace. This does not assert that an arbitrary eigenvector of $T$ is already an eigenvector of $S$.

On each nonzero eigenspace $V_{\lambda }$, the restricted linear map $S_{\lambda }=S{|}_{V_{\lambda }}$ is diagonalizable. Indeed $m_S(S_{\lambda })=0$, so the minimal polynomial of $S_{\lambda }$ divides $m_S$. A divisor of a product of distinct linear polynomials has the same property, proving the diagonalizable restriction lemma.

Choose a basis of $S_{\lambda }$-eigenvectors in each $V_{\lambda }$. Every chosen vector is simultaneously a $T$-eigenvector, with eigenvalue $\lambda$. The direct sum in part (ii) shows that joining these bases gives a basis of $V$. Therefore $S$ and $T$ have a common eigenbasis, which is exactly simultaneous diagonalization.

Suppose a single basis gives the diagonal matrices $D_S,D_T$. Entrywise multiplication gives $D_S{D}_T=D_T{D}_S$. If $P$ is the change-of-basis matrix, thenThus simultaneous diagonalization implies commutation.