2
$\begingroup$

I am sorry if this is stupid in advance. I know how to find eigenvalues of a matrix, but I don't understand how I am suppose to know in which order I am supposed to apply them, in order to find eigenvectors?

Example:

$$ A=\begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \\ \end{pmatrix} $$ Which has eigenvalues of -5,-5,1.

Eigenvector of -5 is:

$$ E(-5)=y\begin{pmatrix} -1\\ 1 \\ 0 \\ \end{pmatrix}+z\begin{pmatrix} -1\\ 0 \\ 1 \\ \end{pmatrix} $$

Eigenvector of 1 is $$ E(1)=y\begin{pmatrix} 1\\ 1 \\ 1 \\ \end{pmatrix} $$

So basically that's ok. Now, how I am supposed in which order I should reorganize these eigenvectors? I mean, the P could be: $$ P_1=\begin{pmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix}=E(-5)+E(1) $$

But it also could be $$ P_2=\begin{pmatrix} 1 & -1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}=E(1)+E(-5) $$ But the $P_1^{-1}$ and $P_2^{-1}$ are different! Thus, $P_2^{-1}AP$ doesn't give an diagonalized matrix, but $P_1^{-1}AP$ does.

What did I miss? How I can be sure to rearrange eigenvector in right order before proceeding to calculate $P^{-1}$?

EDIT Ok thank you very much, I should have made some mistake when calculating one of inverse matrices. I understood that rearranging eigen(values/vectors) doesn't change the fact that I will still must get a diagonalized matrix at the end.

$\endgroup$
  • $\begingroup$ So often what students think is "stupid" might just as well be called "related to learning". $\endgroup$ – The Count May 15 '17 at 21:50
  • $\begingroup$ As long as the eigenvalues and eigenvectors line up, you get a diagonal matrix (no one said that a diagonalization is unique). $\endgroup$ – Michael Burr May 15 '17 at 21:54
2
$\begingroup$

$P_1^{-1} A P_1 = \begin{bmatrix} -5\\&-5\\&&1 \end{bmatrix}\\ P_2^{-1} A P_2 = \begin{bmatrix} 1\\&-5\\&&-5 \end{bmatrix}$

Here is what I think about.

$A\mathbf v_1 = \lambda_1\mathbf v_1\\ A\begin{bmatrix} \mathbf v_1,\mathbf v_2,\mathbf v_3\end{bmatrix} = \begin{bmatrix} \mathbf v_1,\mathbf v_2,\mathbf v_3\end{bmatrix}\begin{bmatrix} \lambda_1\\&\lambda_2\\&&\lambda_3\end{bmatrix}$

$\endgroup$
1
$\begingroup$

$P_1^{-1}AP_1$ are $P_2^{-1}AP_2$ are both diagonal matrices.

You are just reordering the eigenvalues for the diagonal matrices.

$$P_1^{-1}AP_1 = diag(-5, -5, 1) $$

$$P_2^{-1}AP_2 = diag(1, -5, -5) $$

If $Av_j = \lambda_j v_j$, then

$$A \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix} = \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix}diag(\lambda_1 , \ldots , \lambda_n) $$

Let $P = \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix}$ and $D= diag( \lambda_1 , \ldots , \lambda_n)$,

then we have $$AP=PD$$ and hence

$$P^{-1}AP=D$$

Note that the order of $v_i$ controls the order of $\lambda_i$ in the diagonal matrix.

$\endgroup$
  • $\begingroup$ So actually I shoud have a mistake when calculating $$P_2^{-1} $$, right? Reordering eigenvalues doesn't change the fact that I shoud find a diagonalized matrix at the end? $\endgroup$ – dgan May 15 '17 at 21:58
  • $\begingroup$ yup, theoretically, you should still get a diagonal matrix. The error could be in $P_2^{-1}$ or other calculations. $\endgroup$ – Siong Thye Goh May 15 '17 at 21:58

Your Answer

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

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