1
$\begingroup$

I'm trying to find the diagonalization of a matrix :

this is my matrix :

$$ A =\begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix} $$

for the eigenvalues I found :

$$l_1 = l_2 = 1\quad; \quad l_3=-1$$

and for the eigenvectors I found :

$$ v_1 =\begin{pmatrix} -1\\-1\\1 \end{pmatrix} $$ $$v_2=\begin{pmatrix} 0\\0\\1 \end{pmatrix}\quad\text{and}\quad v_3 =\begin{pmatrix} -1\\1\\0 \end{pmatrix} $$

but if I want to find the matrice P will it be :

$$ A =\begin{pmatrix} 0 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \\ \end{pmatrix} $$

Or:

$$ A =\begin{pmatrix} -1 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & -1 \\ \end{pmatrix} $$

Or:

$$ A =\begin{pmatrix} 1 & 0 & -1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \\ \end{pmatrix} $$

etc....

I mean there is many ways to write the P matrix, but which one is the correct one ?

$\endgroup$
  • 1
    $\begingroup$ They're all correct. Every $P^{-1}AP$ will be diagonal. But the diagonal elements (= eigenvalues) are shuffled from one to the other. $\endgroup$ – Julien Apr 29 '13 at 20:27
  • $\begingroup$ @Amzoti That's too kind, but thanks! $\endgroup$ – Julien Apr 30 '13 at 3:02
1
$\begingroup$

It depends on how you write the diagonal matrix. Let's assume that $ D=P^-1AP $. Let's assume that we have the following eigenvalues (they are not necessarily different), $ \lambda_1 ,...\lambda_n $, with the following eigenvectors $ \forall i \in [1,n], Av_i=\lambda_i v_i$.

If we write $ D=diag(\lambda_1....\lambda_n)$, then $ P=(v_1....v_n)$, meaning, we put the eigenvector in the column of the fitting eigenvalue in the diagonal matrix.

$\endgroup$

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.