__Statement__:

__Statement__:

__Proof__:

__Proof__

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Friday, February 3, 2023

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.

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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