Jordan Canonical Form

Let A be a square matrix defined by $ A = \begin{pmatrix}3&2\\ 3&-2\end{pmatrix} $

(a) Find the characteristic polynomial of A.

(b) Show that A is diagonalizable then diagonalize it.

Solutions

Expert Solution

Solution

(a) Find the characteristic polynomial of A.

$ \implies P(\lambda)=|A-\lambda I|=\lambda^2+\lambda-12=\bigg(\lambda-4\bigg)\bigg(\lambda+3\lambda\bigg) $

Therefore, the characteristics polynomial is

$ P(\lambda)=|A-\lambda I|=\bigg(\lambda-4\bigg)\bigg(\lambda+3\lambda\bigg) $

(b) Show that A is diagonalizable then diagonalize it.

since, $ P(\lambda)\hspace{2mm}splitted \implies A $ is diagonalizable.

$ \bullet\hspace{2mm} For \hspace{2mm}\lambda_1=4 $ $ \implies E_4=span\left\{\begin{pmatrix}2\\ 1\end{pmatrix}\right\} $

$ \bullet\hspace{2mm} For\hspace{2mm}\lambda_2=-3 $ $ \implies E_{-3}=span\left\{\begin{pmatrix}-1\\ 3\end{pmatrix}\right\} $

Therefore, $ A=PDP^{-1} $ where

$ P=\begin{pmatrix}2&-3\\ 1&3\end{pmatrix} ,D=\begin{pmatrix}4&0\\ 0&-3\end{pmatrix} ,P^{-1}=\begin{pmatrix}\frac{3}{7}&\frac{1}{7}\\ -\frac{1}{7}&\frac{2}{7}\end{pmatrix} $

since. $ A=PDP^{-1} \implies A^n=\bigg(PDP^{-1}\bigg)^n=PD^nP^{-1} $

$ \implies A^n=\begin{pmatrix}2&-3\\ 1&3\end{pmatrix}\begin{pmatrix}4^n&0\\ 0&\left(-3\right)^n\end{pmatrix}\begin{pmatrix}\frac{3}{7}&\frac{1}{7}\\ -\frac{1}{7}&\frac{2}{7}\end{pmatrix} $

$ =\begin{pmatrix}\frac{3\cdot \:2^{2n+1}+3\left(-3\right)^n}{7}&\frac{2^{2n+1}-6\left(-3\right)^n}{7}\\ \frac{3\cdot \:4^n-3\left(-3\right)^n}{7}&\frac{4^n+6\left(-3\right)^n}{7}\end{pmatrix} $

Therefore, $ A^n=\begin{pmatrix}\frac{3\cdot \:2^{2n+1}+3\left(-3\right)^n}{7}&\frac{2^{2n+1}-6\left(-3\right)^n}{7}\\ \frac{3\cdot \:4^n-3\left(-3\right)^n}{7}&\frac{4^n+6\left(-3\right)^n}{7}\end{pmatrix} $

Answer

Therefore.

a). $ P(\lambda)=|A-\lambda I|=\bigg(\lambda-4\bigg)\bigg(\lambda+3\lambda\bigg) $

b). $ A=PDP^{-1} $ where $ P=\begin{pmatrix}2&-3\\ 1&3\end{pmatrix} ,D=\begin{pmatrix}4&0\\ 0&-3\end{pmatrix} $,$ P^{-1}=\begin{pmatrix}\frac{3}{7}&\frac{1}{7}\\ -\frac{1}{7}&\frac{2}{7}\end{pmatrix} $

c). $ A^n=\begin{pmatrix}\frac{3\cdot \:2^{2n+1}+3\left(-3\right)^n}{7}&\frac{2^{2n+1}-6\left(-3\right)^n}{7}\\ \frac{3\cdot \:4^n-3\left(-3\right)^n}{7}&\frac{4^n+6\left(-3\right)^n}{7}\end{pmatrix} $

SUN BUNRA answered 2 years ago

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