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

A linear endomorphism is diagonalizable over its field precisely when its minimal polynomial is a product of distinct linear polynomials over that same field. Splitting alone is insufficient: repeated factors permit nontrivial Jordan blocks.

For the displayed matrix, a determinant expansion givesThese formulas can be reduced in each requested characteristic. The four cases below distinguish failure to split from failure to have distinct factors.

Over the complex numbers, the three eigenvalues areThey are distinct. Each is a root of the minimal polynomial, so here the minimal polynomial equals the characteristic polynomial and has distinct linear polynomials. Consequently $A$ is diagonalizable over $\mathbb{C}$.

The quadratic polynomial $x^2+4x+10=(x+2{)}^2+6$ has no real roots of a polynomial. Thus the characteristic polynomial does not split over the real numbers. A real diagonal matrix would have only real eigenvalues, and similar matrices have the same characteristic polynomial. Therefore $A$ is not diagonalizable over $\mathbb{R}$.

In the finite field ${\mathbb{F}}_3$, reduce the characteristic polynomial to $(x-1{)}^3$. For $B=A-I$, reduction of the entries givesHenceIts repeated linear polynomial prevents diagonalizability. Equivalently, a diagonalizable map with only the eigenvalue one would have to be $I$, and this matrix is not $I$.

In the finite field ${\mathbb{F}}_2$, the matrix becomesOn the first two coordinate vectors it is a nonzero nilpotent Jordan block of size two, and on the last it is the identity map. ConsequentlyThe repeated zero root prevents diagonalizability. Directly, the only possible eigenvalues are zero and one, so diagonalizability would imply $A^2=A$, whereas the displayed off-diagonal entry disappears on squaring.

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.

Yes, provided “commute” means pairwise commute. Necessity follows because three diagonal matrices commute pairwise.

For sufficiency, first choose the common eigenbasis for the first two linear maps, say $S,T$, obtained in part (b). Their joint eigenspacesgive a direct sum decomposition of $V$. The third map $U$ preserves each $W_{\lambda ,\mu }$, since $SUv=USv=\lambda Uv$ and $TUv=UTv=\mu Uv$. By the diagonalizable restriction lemma, $U{|}_{W_{\lambda ,\mu }}$ has a basis of eigenvectors there. Every such vector already has fixed eigenvalues for $S$ and $T$.

Joining these bases therefore gives a common eigenbasis for all three maps. The same induction proves simultaneous diagonalization for any finite pairwise commuting family of diagonalizable linear endomorphisms.