Question

In: Advanced Math

Jordan Fanonical Form

Let A be a square matrix defined by $ A =\begin{pmatrix}-1&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix} $ and its characteristics polynomial $ P(\lambda)=\bigg(\lambda+1\bigg)^2\bigg(\lambda-2\bigg)^2 $

(a) Find the minimal polynomial of A.

(b) Deduce that A is not diagonalizable, but it is triangularizable, then triangularize A.

Solutions

Expert Solution

Solution

(a) Find the minimal polynomial of A.

we have $ A =\begin{pmatrix}-1&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix} $ and $ P(\lambda)=\bigg(\lambda+1\bigg)^2\bigg(\lambda-2\bigg)^2 $

$ \implies m_A({\lambda})=\bigg(\lambda+1\bigg)^{\alpha_1}\bigg(\lambda-2\bigg)^{\alpha_2} \hspace{2mm} $ where $ \alpha_1,\alpha_2 $ are index of

$ \lambda_1,\lambda_2 $ respectivdy.

$ A+I=\begin{pmatrix}0&-2&-1&3\\ -6&-5&1&6\\ -6&-4&0&6\\ -6&-7&1&8\end{pmatrix}\sim \begin{pmatrix}0&-2&-1&3\\ -6&-4&1&6\\ 0&0&0&0\\ 0&-3&0&3\end{pmatrix}\sim \begin{pmatrix}0&-2&-1&3\\ -6&0&3&0\\ 0&0&0&0\\ 0&-3&0&3\end{pmatrix} $

Let $ x_4=t\implies x_2=t, x_3=S\implies x_1=-2x_2+3x_3 ,x_1=\frac{t}{2} $

Then $ E_{\lambda_1}=span\left\{\begin{pmatrix}\frac{1}{2}\\ 1\\ 1\\ 1\end{pmatrix}\right\} \iff gm(\lambda_1)=1\neq am(\lambda_1)\implies \alpha_1=2 $

$ \implies A-2I=\begin{pmatrix}-3&-2&-1&3\\ -6&-7&1&6\\ -6&-4&-2&6\\ -6&-7&1&6\end{pmatrix}\sim \begin{pmatrix}-3&-2&-1&3\\ -6&-7&1&6\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}\sim \begin{pmatrix}-3&-2&1&3\\ 0&-3&3&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} $

Let $ x_3=t\hspace{2mm},x_2=t\hspace{2mm}and\hspace{2mm} x_4=S\implies x_1=S-\frac{t}{3} $

Then $ E_{\lambda_2}=span\left\{\begin{pmatrix}1\\ 0\\ 0\\ 1\end{pmatrix},\begin{pmatrix}-\frac{1}{3}\\ 1\\ 1\\ 0\end{pmatrix}\right\}\implies gm(\lambda_2)=am(\lambda_2)=2\implies \alpha_2=2 $

Thus, $ m_A(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda+1\bigg)^2 $

(b) Deduce that A is not diagonalizable, but it is triangularizable, then triangularize A.

since $ am(\lambda_1)\neq gm(\lambda_1) $

so A is not diagonalizable. But it's triangularizable

$ \bullet \hspace{2mm} Tringularizable \hspace{2mm}that\hspace{2mm} is \hspace{2mm} A=PJP^{-1} $

where $ J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&1\\ 0&0&0&-1\end{pmatrix}\hspace{2mm}, $

$ \bigg(A+I\bigg)^2=\begin{pmatrix}0&-9&0&-9\\ -18&-18&9&18\\ -18&-18&8&18\\ -18&-27&9&27\end{pmatrix}\sim \begin{pmatrix}0&1&0&1\\ 2&2&-1&-2\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix} $

Let $ x_4=t\implies x_2=-t\hspace{2mm},x_3=S\implies x_1=\frac{1}{2}S+2t $

Thus, $ E_(\lambda_1)^2=span\left\{\begin{pmatrix}\frac{1}{2}\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}2\\ -1\\ 0\\ 1\end{pmatrix}\right\} $

Let $ V_2\in E_{\lambda_1}^2-E_{\lambda_1}\implies V_2=\begin{pmatrix}\frac{1}{2}\\ 0\\ 1\\ 0\end{pmatrix} $

$ \implies V_1=\bigg(A+I\bigg)V_2=\begin{pmatrix}0&-2&-1&3\\ -6&-4&1&6\\ -6&-4&1&6\\ -6&-7&1&9\end{pmatrix}\sim \begin{pmatrix}\frac{1}{2}\\ \:0\\ \:1\\ \:0\end{pmatrix}=\begin{pmatrix}-1\\ -2\\ -2\\ -2\end{pmatrix} $

Then, $ P\bigg(u_1,u_2,V_1,V_2\bigg)=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} $

Thus, $ A=PJP^{-1}\hspace{2mm},J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix},P=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} $

$ \implies A=PJP^{-1}\iff A^n=PJ^nP^{-1}\hspace{2mm},J=D+M $

since $ D=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix} $ and $ M=\begin{pmatrix}0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}\hspace{2mm} where \hspace{2mm}M^2=0 $

so,$ J^n=D^n+nD^{n-1}M=\begin{pmatrix}2^n&0&0&0\\ 0&2^n&0&0\\ 0&0&\left(-1\right)^n&n\left(-1\right)^{n-1}\\ 0&0&0&\left(-1\right)^{n\:}\end{pmatrix} $

Answer

Therefore

a). $ m_A(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda+1\bigg)^2 $

b). $ A=PJP^{-1}\hspace{2mm},J=\begin{pmatrix}2&0&0&0\\ 0&2&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix},P=\begin{pmatrix}1&-\frac{1}{3}&-1&\frac{1}{2}\\ 0&1&-2&0\\ 0&1&-2&1\\ 1&0&-2&0\end{pmatrix} $

c). $ A^n=P\begin{pmatrix}2^n&0&0&0\\ \:0&2^n&0&0\\ \:0&0&\left(-1\right)^n&n\left(-1\right)^{n-1}\\ \:0&0&0&\left(-1\right)^{n\:}\end{pmatrix}P^{-1} $

SUN BUNRA answered 2 years ago

Next > < Previous

Related Solutions

Jordan Canonical Form

Let $ A\in M_n(\mathbb{R})\hspace{2mm} $ and $ \hspace{2mm}\lambda_1, \lambda_2,...,\lambda_n \hspace{2mm} $(no need distinct) be eigenvalues of A. Show that a). $ \sum _{i=1}^n\lambda _i=tr\left(A\right) $ b). $ \:\prod _{i=1}^n\lambda _i=\left|A\right|\: $

Jordan Canonical Form

Let $ A\in M_n(\mathbb{R})\hspace{2mm} $ and$ \hspace{2mm} m_A(\lambda)\hspace{2mm} $ be its minimal polynomial. Let f be a polynomial satisfies$ \hspace{2mm}f(A) = 0. \hspace{2mm} $Show that$ \hspace{2mm} f(\lambda) \hspace{2mm} $is divisible by$ \hspace{2mm} m_A(\lambda). $

Jordan Canonical Form

Let A be a square matrix defined by $ A=\begin{pmatrix}4&-2&1\\ 2&0&1\\ 2&-2&3\end{pmatrix}\hspace{2mm} $Find the minimal polynomial of A. Then express $ A^4 $ and $ A^{-1} $ in terms of A and I.

Jordan Canonical Form

Determine the value of a so that $ \lambda = 2 $ is an eigenvalue of $ A=\begin{pmatrix}1&-1&0\\ a&1&1\\ 0&1+a&3\end{pmatrix} $ Then show that A is diagonallizable and diagonalize it.

Jordan Canonical Form

Let $ A\in M_6(\mathbb{R}) $ be an invertible matrix satisfies $ A^3-4A^2 + 3A = 0 $ and $ tr(A) = 8. $ Find the characteristics polynomial of A.

Jordan Canonical Form

Let A be a square matrix defined by $ A = \begin{pmatrix}-3&-1&-3\\ 5&2&5\\ -1&-1&-1\end{pmatrix} $ (a) Find the characteristic polynomial of A. (b) Find the eigenvalues of A. Show that A is not diagonalizable over $ \mathbb{R} $ (c) Show that A is diagonalizable over$ \mathbb{C} $. Find the eigenspaces. Diagonalize A. (d) Express $ A^n $ in the form of $ a_nA^2+b_ nA+c_nI_n $ where $ (a_n), (b_n) $ and $ (c_n) $ are real sequences to be specified....

Jordan Canonical Form

Let A be a square matrix defined by $ A = \begin{pmatrix}-2&-1&-5\\ 2&2&3\\ 4&2&2\end{pmatrix} $ (a) Find the characteristic polynomial of A. (b) Find the eigenvalues and eigenspaces of A. (c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A. (d) Find the three real sequences $ (a)_n, (b)_n ,(c)_n $ satisfying. $ \begin{cases} a_{n+1}=-2a_n-b_n-5c_n \hspace{2mm},a_0=1 & \quad \\ b_{n+1}=2a_n+2b_n+3c_n \hspace{2mm}, b_0=0 & \quad \\ c_{n+1}=4a_n+2b_n+6c_n \hspace{2mm},c_0=1 & \quad \end{cases} $

Jordan Canonical Form

Let A be a square matrix defined by $ A =\begin{pmatrix}-8&-3&-6\\ 4&0&4\\ 4&2&2\end{pmatrix} $ (a) Find the characteristic polynomial of A. (b) Find the eigenvalues and eigenspaces of A. (c) Show that A is not diagonalizable, but it is triangularizable, then triangularize A. (d) Write $ A^n $ in terms of $ I, A,A^2 $ and n.

Jordan Canonical Form

Let A be a square matrix defined by$ A =\begin{pmatrix}2&-3&1\\ 1&-2&1\\ 1&-3&2\end{pmatrix} $ (a) Find the characteristic polynomial of A. (b) Show that A is diagonalizable then diagonalize it. (c) Write $A^n$ \hspace{2mm} in term of n.

Jordan Canonical Form

Let A be a square matrix defined by $ A =\begin{pmatrix}-1&3&-1\\ -3&5&-1\\ -3&3&1\end{pmatrix} $ (a) Find the characteristic polynomial of A. (b) Show that A is diagonalizable then diagonalize it. (c) Write $ A^n $ in term of n.

Subjects