Diagonalisation and the Minimal Polynomial: Examples and Counterexamples

On the zero space, $I=0$, so $1$ annihilates and [L1] gives $\mu =1$. On a nonzero space, $(\lambda I-\lambda I)=0$, while no nonzero constant polynomial annihilates; hence ${\mu }_{\lambda I}=x-\lambda$.

For the diagonal matrix, a polynomial vanishes on the matrix exactly when it vanishes at every displayed diagonal value. The monic polynomial of least degree with those distinct roots is ${\prod }_i(x-{\lambda }_i)$, agreeing with [L2].

By [L1], both characteristic polynomials are $(x-1{)}^2$.

The polynomial $x-1$ annihilates $I$. For $N=J-I\mathrm{\ne }0$, one has $N^2=0$, so $(J-I{)}^2=0$ but $J-I\mathrm{\ne }0$. Hence [L2] gives ${\mu }_I=x-1$ and ${\mu }_J=(x-1{)}^2$.

Direct multiplication gives $A^2=A$, while $A\mathrm{\ne }0$ and $A\mathrm{\ne }I$. Thus $x(x-1)$ annihilates but neither $x$ nor $x-1$ does; [L1] gives ${\mu }_A=x(x-1)$.

Expanding $\det (xI-A)$ gives $x(x-1{)}^2$ by [L3]. Finally [L2] identifies the image-kernel projection.

For $n>0$, repeated application of $T$ gives $T^n=0$, while $T^{n-1}{e}_{n-1}=e_0\mathrm{\ne }0$ by [L1]. Hence [L2] gives ${\mu }_T=x^n$.

The sole irreducible factor is $x$, so [L3] gives the one primary component $\ker T^n=F^n$. At $n=0$, [L1] and [L2] give ${\mu }_T=1$.

Direct multiplication gives $R^2=-I$, while no linear polynomial annihilates $R$, so ${\mu }_R=x^2+1$. By [L2] and [L1], $R$ is not diagonalisable over $\mathbb{R}$.

Over $\mathbb{C}$, [L3] gives $x^2+1=(x-i)(x+i)$ with distinct roots. By [L1] the extended matrix is diagonalisable; explicitly $(i,1)$ and $(-i,1)$ are eigenvectors for $i$ and $-i$ and form a basis.

The first block has minimal polynomial $x-2$, and $C^2=-I$ makes the second block's minimal polynomial $x^2+1$; [L3] rules out a rational linear factor. The two factors are coprime, so the direct sum has minimal polynomial their product and [L1] identifies the two primary components.

The polynomial $e_1=(x^2+1)\mathrm{/}5$ is $1$ at $x=2$ and is $0$ modulo $x^2+1$; $e_2=1-e_1=(4-x^2)\mathrm{/}5$ has the opposite residues. Therefore [L2] gives the displayed projections.

The denominators are nonzero, and substitution gives $e_i({\lambda }_j)=1$ for $i=j$ and $0$ otherwise. Hence $e_i(T)$ is identity on $E_{{\lambda }_i}(T)$ and zero on the other eigenspaces.

The action on every summand in [L1] now gives $E_i^2=E_i$, $E_i{E}_j=0$ for $i\mathrm{\ne }j$, and ${\sum }_i{E}_i=I$.

With $P^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$, direct multiplication gives $A=P\mathrm{diag}(1,2)P^{-1}$ and $B=P\mathrm{diag}(3,4)P^{-1}$.

Thus the columns of $P$, namely $(1,0)$ and $(1,1)$, are a common eigenbasis. The two diagonal matrices commute, so their conjugates commute as well, agreeing with [L1].

The identity commutes with $J$, so the family is pairwise commuting.

Since $(J-I{)}^2=0$ but $J-I\mathrm{\ne }0$, the minimal polynomial of $J$ is $(x-1{)}^2$. Thus [L2] says $J$ is not diagonalisable, so the family cannot be simultaneously diagonalisable. The missing hypothesis in [L1] is diagonalisability of every member.

One has $(J-I{)}^2=0$ but $J-I\mathrm{\ne }0$, so ${\mu }_J=(x-1{)}^2$. This polynomial splits by [L2], but its root is repeated.

By [L1], $J$ is not diagonalisable. Thus splitting alone is insufficient; squarefreeness is the missing condition.

By [L1], both matrices have characteristic polynomial $(x-1{)}^2$.

The identity is diagonal. For $J$, $(J-I{)}^2=0$ but $J-I\mathrm{\ne }0$, so ${\mu }_J=(x-1{)}^2$ and [L2] says $J$ is not diagonalisable. Thus the shared characteristic polynomial does not decide diagonalisability.

Both standard basis vectors are eigenvectors of $I_2$ with eigenvalue $1$, so [L1] and [L2] make $I_2$ diagonalisable.

Its matrix is diagonal with two entries $1$, so [L3] gives ${\chi }_{I_2}=(x-1{)}^2$, which has a repeated root.