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Thank you ahead of time for the help, I am having a problem with part $4$. I understand parts $1$ and $2$ and $3$ and have solved them but I cant seem to understand $4$. If someone could help me out, that would be amazing.

Let $A = \begin{pmatrix}0 & -4 & -6\\-1 & 0 & -3\\1 & 2 & 5\end{pmatrix}$

  1. Find the characteristic polynomial and the eigenvalues of $A$.
  2. Find a basis for each eigenspace.
  3. Is $A$ diagonalizable? If yes, diagonalize $A$.
  4. Find $A^{10}$

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2 Answers 2

up vote 3 down vote accepted

Hint: If you know the characteristic polynomial, then express $x^{10}=p(x)q(x)+r(x)$ where $q(x)$ is the characteristic polynomial and $r(x)=0$ or,degree of $r(x)$ is less than or equal 2. Then by Cayley-Hamilton theorem, $A^{10}=r(A)$.

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not sure exactly what you mean –  user161059 Jul 1 at 12:05
    
@ user161059. Okay, Please can you type the characteristic polynomial. I am sorry I didn't compute it –  Chellapillai Jul 1 at 12:09
    
Also we can play with the minimal polynomial $(x-5)(X-4)$ if computation is relatively fast-track. –  Chellapillai Jul 1 at 12:46

You just have to do the following:

$A^{10}=(S^{-1}DS)^{10}=S^{-1}DS \cdot S^{-1}DS\cdot....\cdot S^{-1}DS=S^{-1}D^{10}S$ (Because $S^{-1}S=I$)

For your diagonalizable matrix A and the diagonal matrix D

Edit:

For more information see http://en.wikipedia.org/wiki/Matrix_diagonalization#Diagonalization

http://www.wolframalpha.com/input/?i=diagonalize+%7B%7B0%2C-4%2C-6%7D%2C%7B-1%2C0%2C-3%7D%2C%7B1%2C2%2C5%7D%7D

Note: In the second link you can see that your eigenvalues are actually wrong. The eigenvalues of the matrix are $\lambda_1=1$ and $\lambda_{2,3}=2$

If you don't know how to compute them, ask a new question.

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for one of the questions i got D= 4 0 0, 0 5 0, 0 0 5, how would i do this to a 10th power...how would i approach it... –  user161059 Jul 1 at 11:48
    
Just multiply two diagonal matrices and you will see that the rule for diagonal matrices is: If $D=( \lambda_1, \lambda_2,..., \lambda_m)$ is a diagonal matrix, then $D^n=( \lambda_1^n, \lambda_2^n,..., \lambda_m^n)$ –  Mesih Jul 1 at 11:51
1  
$D^{10}$=\begin{pmatrix} 4^{10} & 0 & 0 \\ 0 & 5^{10} & 0 \\ 0 & 0 & 5^{10} \end{pmatrix} –  Mesih Jul 1 at 12:05
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Compute $S^{-1}D^{10}S$ and you are done. –  Mesih Jul 1 at 12:08
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S is the transformation matrix, such that $S^{-1}AS=D$. D is your diagonal matrix which you computed before (it contains your eigenvalues, they have to be linear independent!) and S is the matrix which contains three different eigenvectors in its column. Well, above i wrought $SAS^{-1}=D$ in that case the matrix which contains the eigenvectors is $S^{-1}$. It is important that in the formula $S^{-1}AS=D$ the matrix on the right side is the matrix which contains the eigenvectors and the matrix on the left side is its inverse. I edited a link above. –  Mesih Jul 1 at 12:19

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