# Cayley Hamilton Theorem

### Statement:

Let T be a linear operator on a finite dimensional vector space V. If  f is the characteristic polynomial for T, Then $f(T)=0$, i.e. the minimal polynomial divides the characteristic polynomial.

### Proof:

Let K be the commutative ring with identity consisting of all polynomials in T.
Choose an ordered basis $\{\alpha_1, \alpha_2, ... ,\alpha_n\}$ for V.
Let A be the matrix of T in the basis $\{\alpha_1, \alpha_2, ... ,\alpha_n\}$.
Then we have,
$T(\alpha_i)=\displaystyle\sum_{j=1}^{n}A_{ji}\alpha_j$ ,                              $(1\le i\le n)$
$\implies \displaystyle\sum_{j=1}^{n}(\delta_{ij}T-A_{ji}I)\alpha_j=0$ ,      $(1\le i\le n)$ $........ (1)$
Let $B\in K^{n\times n}$ and the $(i, j)^{th}$ entry of B is given by $B_{ij}=(\delta_{ij}T-A_{ji}I)$.
When $n=2$,        $B = \begin{bmatrix} T-A_{11}I & -A_{21}I \\[0.3em] -A_{12}I & T-A_{22}I \\[0.3em] \end{bmatrix}$
and,
det $B=(T-A_{11}I)(T-A_{22}I)-A_{12}A_{21}I$
$=T^2-(A_{11}+A_{22})T+(A_{11}A_{22}-A_{12}A_{21})I$
$=T^2-$ (trace of A) $T+$ (det $A$)$I$
det $B=f(T)$
where $f$ is a characteristic polynomial and  $f=x^2-(trace \ of \ A) x+(det \ A)$
In case of $n>2$,
det $B=f(T)$ where $f$ is the determinant of the matrix $(xI-A)$ where the $(i, j)^{th}$ entry is $(xI-A)_{ij}=(\delta_{ij}x-A_{ji})$
To show $f(T)=0$ , i.e. to prove $f(T)$ to be a zero operator, it is sufficient to prove that $(det \ B)\alpha_k=0$       for      $(1\le k\le n)$
Now from (1),
$\displaystyle\sum_{j=1}^{n}B_{ij}\alpha_j=0$         ............ (2)
When $n=2$,
$\begin{bmatrix} T-A_{11}I & -A_{21}I \\[0.3em] -A_{12}I & T-A_{22}I \\[0.3em] \end{bmatrix} \begin{bmatrix} \alpha_1 \\[0.3em] \alpha_2 \\[0.3em] \end{bmatrix} = \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix}$
Denote the classical adjoint of $B$ by $\bar{B}$ then
$\bar{B} = \begin{bmatrix} T-A_{22}I & A_{21}I \\[0.3em] A_{12}I & T-A_{11}I \\[0.3em] \end{bmatrix}$
$\therefore \bar{B}B = \begin{bmatrix} T-A_{22}I & A_{21}I \\[0.3em] A_{12}I & T-A_{11}I \\[0.3em] \end{bmatrix} \begin{bmatrix} T-A_{11}I & -A_{21}I \\[0.3em] -A_{12}I & T-A_{22}I \\[0.3em] \end{bmatrix}$
$=\begin{bmatrix} (T-A_{11}I)(T-A_{22}I)-A_{12}A_{21}I & (T-A_{11}I)A_{21}I-A_{21}I(T-A_{11}I) \\[0.3em] -A_{12}I(T-A_{22}I)+(T-A_{22}I)A_{12}I & -A_{12}A_{21}I+(T-A_{11}I)(T-A_{22}I)\\[0.3em] \end{bmatrix}$
$=\begin{bmatrix} det B & 0 \\[0.3em] 0 & det B \\[0.3em] \end{bmatrix}$
$=(det \ B)I$
Hence,
$(det B)\begin{bmatrix} \alpha_1 \\[0.1em] \alpha_2 \\[0.1em] \end{bmatrix}=(det B)I\begin{bmatrix} \alpha_1 \\[0.1em] \alpha_2 \\[0.1em] \end{bmatrix}$
$=(\bar{B}B)\begin{bmatrix} \alpha_1 \\[0.3em] \alpha_2 \\[0.3em] \end{bmatrix}$
$=\bar{B}\Bigg(B\begin{bmatrix} \alpha_1 \\[0.3em] \alpha_2 \\[0.3em] \end{bmatrix}\Bigg)$
$=\bar{B}(0)$
Then $(det \ B)\begin{bmatrix} \alpha_1 \\[0.3em] \alpha_2 \\[0.3em] \end{bmatrix}=0 \implies f(T)=0$
If $n>2$ (in general), denote adj. $B=\bar{B}$.
Then from (2),
$\displaystyle\sum_{j=1}^{n}(\bar{B}_{ki}B_{ij})\alpha_j=0$
For each pair k, i and summing for i, we have,
$0=\displaystyle\sum_{i=1}^{n}\Big(\displaystyle\sum_{j=1}^{n}\bar{B}_{ki}B_{ij}\Big)\alpha_j$
$=\displaystyle\sum_{j=1}^{n}\Big(\displaystyle\sum_{i=1}^{n}\bar{B}_{ki}B_{ij}\Big)\alpha_j$
But, $\Big(\displaystyle\sum_{i=1}^{n}\bar{B}_{ki}B_{ij}\Big)=\delta_{kj}(det \ B)$
Therefore,
$0=\displaystyle\sum_{j=1}^{n}\delta_{kj}(det \ B)\alpha_j=(det \ B)\alpha_k$ ,
$\implies (det \ B)=f(T)=0$
Then if f is a characteristic polynomial of T, then $f(T)=0$.