0
$\begingroup$

Determine whether $A$ is diagonalizable and if it is, calculate $A^5$.

$$ A= \begin{bmatrix} 3 & -1 & -1 \\ -12 & 0 & 5 \\ 4 & -2 & -1 \\ \end{bmatrix} $$

So I calculated the eigenvalues by factoring and got $(\lambda -2)(1- \lambda ^2)=0 $ so $\lambda = -1,1,2$. So I tried to calculate the eigenvectors.

For $\lambda = 1$ I plugged in the eigenvalue to get $$ A- \lambda I= \begin{bmatrix} 2 & -1 & -1 \\ -12 & -1 & 5 \\ 4 & -2 & -2 \\ \end{bmatrix} $$ Which could be row reduced to the identity matrix, so I got an eigenvector of $$ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} $$

Using the same process, for the eigenvalue of $\lambda = -1$, I got an eigenvector of $$ \begin{bmatrix} 1 \\ 2\\ 2\\ \end{bmatrix} $$

and for an eigenvalue of $\lambda = 2 $ I got an eigenvector of $$ \begin{bmatrix} -1 \\ 1 \\ -2 \\ \end{bmatrix} $$

So to figure out $A^5$ I am supposed to use the fact that to diagonalize the matrix, I use $S^{-1}AS=D$ where D is the diagonalized matrix. and I am supposed to use the fact that $SD^5S^{-1}=A^5$

So $$S= \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 2 \\ 0 & -2 & 2 \\ \end{bmatrix} $$

and

$$ D= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} $$

However, my problem is that I know the eigenvector for $\lambda = 1 $ is wrong and I cannot figure out what I am doing wrong. Thank you for anyone that helps.

$\endgroup$
  • 2
    $\begingroup$ $A - I$ cannot be reduced to the identity. Go through that row reduction again. $\endgroup$ – Omnomnomnom Dec 7 '14 at 15:18
  • 1
    $\begingroup$ Also, if $A - \lambda I$ can be row-reduced to the identity, then $\lambda$ is not an eigenvalue. $\endgroup$ – Omnomnomnom Dec 7 '14 at 15:19
  • 1
    $\begingroup$ The zero vector is by definition never an eigenvector. $\endgroup$ – Nishant Dec 7 '14 at 15:27
  • 1
    $\begingroup$ don't you see that the third row $A - I$ is twice that of the first row of $A - I.$ that alone tells you $(2, 0, -1)^T$ is an eigenvector corresponding to the eigenvalue $1.$ $\endgroup$ – abel Dec 7 '14 at 15:47
  • 1
    $\begingroup$ i made a mistake in my earlier comment. $(2, 0, -1)^T$ is in the left null space. the vector $(3, -1, 7)$ is in the (right)null space. apologies. $\endgroup$ – abel Dec 7 '14 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.