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 \( f(T)=0 \).
