LADR Factor and Apply the Dependence

An alternative process to Finding Eigenvectors.

Theorem

• Suppose $T:V\to V$ is a linear map of an $n-dim$ $\mathbb{R}$ vector space.
• Suppose $p(T)=Z$ for some polynomial $p\in P_n(\mathbb{R})$
• Then, factors as $Z=T^{n+n}+\cdots +a_1{T}^1+a_{0}I=(T-{\lambda }_{k}I)\dots (T-{\lambda }_{1}I)$
• For any non-zero vector $v\in V$, we have that one of the vectors $(T-\lambda I)v,(T-{\lambda }_{2}I)(T-{\lambda }_{1}I)v,\dots ,(T-{\lambda }_{k-1}I)\dots (T-{\lambda }_{1}I)v$ is an Eigenvector of $T$

Example

Find a dependence of the powers of

$$\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$$

Use this dependence to find eigenvectors and eigenvalues

1. Find a dependence

$$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$ $$T=\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]$$ $$T^2={\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]}^2=\left[\begin{array}{cc}9& 8\\ 16& 17\end{array}\right]=4T+5I$$

2. This gives $T^2=4T+5I⟺T^2-4T+5I=Z$
3. We can think of this as $x^2-4x-5=0⟺(x-5)(x+1)=0$
4. So, $(T-5I)(T+I)=Z$
5. We get $\lambda =5,\lambda =-1$
6. Pick any non-zero vector and compute
7. $0=Z(1,0)=(T-5I)(T+I)(1,0)$
8. We calculate

$$(\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])\left[\begin{array}{c}1\\ 0\end{array}\right]$$ $$=\left[\begin{array}{cc}2& 2\\ 4& 4\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}2\\ 4\end{array}\right]=(2,4)$$

9. Thus, $(1,0)$ is not a $\lambda =-1$ eigenvector
10. We continue, $(T-5I)(2,4)$ with the process, putting the last thing as an input

$$(\left[\begin{array}{cc}1& 2\\ 4& 3\end{array}\right]-5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])\left[\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$