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$$A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 2 \end{pmatrix}$$

I need to diagnolize this matrix, I found that the characeristic polynomial is $x^3-7x^2+15x-9$ and that the eigenvalues are 3,3 and 1.

I also found that the eigen vectors are for $Ker (A-3I) = span\{(0,1,0), (1,0,1)\} $ and $Ker (A-I) = span(1,0,-1)$ So I figured that the inverse matrix P when $P^{-1}AP = D$ is $$P=\begin{pmatrix} 0&1&1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix}$$ (the eigen vectors as columns).

so $$P^{-1} = \begin{pmatrix} 0&1&0 \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{pmatrix}$$

But when multiplying $P^{-1}AP$ i'm not getting a diagonal form, does the order of the columns in $P$ matter? also what an I doing wrong?

A.Γ.
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D_R
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    @T.Bongers how comes its a duplicate? i'm trying to understand what i'm doing wrong I did follow the steps I know. i'm pretty sure it has to do with the order of the eigen vectors or something – D_R Feb 15 '16 at 17:52
  • I entered the matrices into an online calculator and got a diagonal matrix (with appropriate diagonal elements). You probably made an arithmetic error. – Trevor Norton Feb 15 '16 at 17:55
  • Maxima gives Diag(3,3,1). – Bernard Feb 15 '16 at 17:58

1 Answers1

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Something seems to be wrong in your final calculation of $P^{-1}AP$.

Your eigenvectors are correct, and $P$ does contain them.

You can also check that $PP^{-1}=P^{-1}P=I$.

Finally, as Trevor Norton notes, $P^{-1}A P$ is indeed a diagonal matrix. Let me know if you need more detailed steps in calculating this product.

The order of the columns in $P$ is the order of the eigenvectors. This influences the order of the corresponding eigenvalues in the diagonal matrix.

Roland
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