Cayley Hamilton Theorem

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 $f(T)=0$. 

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