= Repeated eigenvalues in the Cayley-Hamilton theorem
{title2=$B=2I+N,\quad N^2=0$}
Take a different real <matrix>,
$$
B=\begin{pmatrix}2&1\\0&2\end{pmatrix}=2I+N,
\qquad N=\begin{pmatrix}0&1\\0&0\end{pmatrix}.
$$
Its only <eigenvalue> is $2$, with <algebraic multiplicity> two. However, the <polynomial> $q(t)=t-2$, which vanishes at that <eigenvalue>, gives $q(B)=N\ne0$. Merely checking values at <eigenvalues> cannot prove a matrix identity. The full <characteristic polynomial> does work:
$$
\chi_B(t)=(t-2)^2,\qquad \chi_B(B)=N^2=0.
$$
The <matrix> is a <Jordan block>, and its <minimal polynomial> is also $(t-2)^2$.
The <binomial theorem> gives, for positive <integers> $n$,
$$
\boxed{B^n=2^nI+n2^{n-1}N.}
$$
Terms containing $N^2$ or a higher <matrix power> vanish. After removing the common growth factor $2^n$, the remaining <linear map> is the <shear mapping> $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single <eigenvalue> controls the exponential scale, while the <nilpotent matrix> adds a factor growing linearly in $n$.
\Image[/cayley-hamilton-jordan-block-shear.png]
{title=Normalized powers of a Jordan block retain a growing shear despite having only one eigenvalue; equal axes scales show the top edge moving by n/2}
In terms of <polynomial interpolation>, reduction modulo $(t-2)^2$ retains both a value and a <derivative>:
$$
f(B)=f(2)I+f'(2)N.
$$
Indeed, <polynomial division> gives $f(t)=f(2)+f'(2)(t-2)+(t-2)^2h(t)$. This is a simple <Hermite interpolation> condition. For $f(t)=t^n$, it gives exactly the displayed <matrix power>. Repeated <eigenvalues> retain information about how a <polynomial> changes, not just where it vanishes.
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