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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 ?

Croviajo
  • 183

1 Answers1

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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.

AnonymouseCat
  • 1,221