Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A^{10}$

share|cite|improve this question
up vote 2 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, $A^{10}=r(A)$.

share|cite|improve this answer
not sure exactly what you mean – user161059 Jul 1 '14 at 12:05
@ user161059. Okay, Please can you type the characteristic polynomial. I am sorry I didn't compute it – Chellapillai Jul 1 '14 at 12:09
Also we can play with the minimal polynomial $(x-5)(X-4)$ if computation is relatively fast-track. – Chellapillai Jul 1 '14 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


For more information see

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.

share|cite|improve this answer
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 would i approach it... – user161059 Jul 1 '14 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 $D^n=( \lambda_1^n, \lambda_2^n,..., \lambda_m^n)$ – Marm Jul 1 '14 at 11:51
$D^{10}$=\begin{pmatrix} 4^{10} & 0 & 0 \\ 0 & 5^{10} & 0 \\ 0 & 0 & 5^{10} \end{pmatrix} – Marm Jul 1 '14 at 12:05
Compute $S^{-1}D^{10}S$ and you are done. – Marm Jul 1 '14 at 12:08
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. – Marm Jul 1 '14 at 12:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.