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I need to find an orthogonal matrix $P$ such that $P^tAP$ is diagonal, where:

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

I know that this matrix $P$ exists because $A^tA=AA^t$ ($A$ is normal).

The characteristic polynomial of $A$ is $p(x)=x^3(x-4)$. The eigenvectors are $v_1=(-1,1,0,0)$, $v_2=(-1,0,1,0)$, $v_3(-1,0,0,1)$ (for $x=0$) and $v_4=(1,1,1,1)$ (for $x=4$).

I know eigenvectors associated with different eigenvalues ​​are orthogonal, but $v_1,v_2,v_3$ are not orthogonal vectors.

I try to find $P$ applying Gram-Schmit procedure with $v_1,v_2,v_3,v_4$ but with this matrix do not hold that $P^tAP$ is diagonal.

Is this idea correct? maybe I'm making a miscalculation to find $P$ or I'm doing something else wrong?

Thanks for your help.

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    I will try to find my error (again). Thanks. – 2012-12-12

2 Answers 2

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Gram-Schmidt should work. :-)

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    Thank you very much! I will try to find my error. By "change the form" I was trying to say that the new vectors (after applying G-S) does not work. but according to your comments, I'm wrong. Again thank you very much :) – 2012-12-12
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Keep $v_1$ and $v_4$ and change $v_2, v_3$ until they are all mutually orthogonal eigenvectors. This can all be done with vectors which have entries from $\{-1,0,1\}$, so there are only finitely many possibilities to try.