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How do you orthogonally diagonalize the matrix A?

Matrix A =

$$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} $$

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  • $\begingroup$ Did you know that eigenvectors of distinct eigenvalues are mutually orthogonal? $\endgroup$ – Easy May 4 '13 at 3:13
  • $\begingroup$ An algorithm is given on the Wikipedia page: en.wikipedia.org/wiki/Orthogonal_diagonalization $\endgroup$ – Potato May 4 '13 at 3:14
  • $\begingroup$ SEE math.stackexchange.com/questions/375711/… $\endgroup$ – Will Jagy May 4 '13 at 3:38
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    $\begingroup$ The same way you orthogonally diagonalize any symmetric matrix: you find the eigenvalues, you find an orthonormal basis for each eigenspace, you use the vectors in the orthogonal bases as columns in the diagonalizing matrix. $\endgroup$ – Gerry Myerson May 4 '13 at 3:54
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Since the matrix $A$ is symmetric, we know that it can be orthogonally diagonalized. We first find its eigenvalues by solving the characteristic equation:

$$0=\det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 1 & 1 \\ 1 & 1-\lambda & 1 \\ 1 & 1 & 1-\lambda \end{vmatrix}=-(\lambda-3)\lambda^2 \implies \left\{\begin{array}{l l} \color{red}{\lambda_1 = 0} \\ \color{green}{\lambda_2 = 0} \\ \color{blue}{\lambda_3 = 3} \end{array}\right.$$

We now find the eigenvectors corresponding to $\lambda=0$:

$$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \end{array}\right) \implies \left(\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \implies \mathbf{x}=\pmatrix{s\\t\\-s-t}=s\pmatrix{1\\0\\-1}+t\pmatrix{0\\1\\-1}$$

By orthonormalizing them, we obtain the basis

$$\left\{\color{red}{\frac{1}{\sqrt{2}}\pmatrix{1\\0\\-1}},\color{green}{\frac{1}{\sqrt{6}}\pmatrix{-1\\2\\-1}}\right\}$$

We finally find the eigenvector corresponding to $\lambda=3$:

$$\left(\begin{array}{ccc|c} -2 & 1 & 1 & 0 \\ 1 & -2 & 1 & 0 \\ 1 & 1 & -2 & 0 \end{array}\right) \implies \left(\begin{array}{ccc|c} 0 & -3 & 3 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 3 & -3 & 0 \end{array}\right) \implies \left(\begin{array}{ccc|c} 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \implies \mathbf{x}=\pmatrix{s\\s\\s}=s\pmatrix{1\\1\\1}$$

By normalizing it, we obtain the basis

$$\left\{\color{blue}{\frac{1}{\sqrt{3}}\pmatrix{1\\1\\1}}\right\}$$

Hence $A$ is orthogonally diagonalized by the orthogonal matrix

$$P=\pmatrix{\color{red}{1/\sqrt{2}}&\color{green}{-1/\sqrt{6}}&\color{blue}{1/\sqrt{3}}\\\color{red}{0}&\color{green}{2/\sqrt{6}}&\color{blue}{1/\sqrt{3}}\\\color{red}{-1/\sqrt{2}}&\color{green}{-1/\sqrt{6}}&\color{blue}{1/\sqrt{3}}}$$

Furthermore,

$$P^{T}AP=\pmatrix{\color{red}{0}&0&0\\0&\color{green}{0}&0\\0&0&\color{blue}{3}}$$

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  • $\begingroup$ Thank you Tharsis! I am now trying your method for the next few questions. Now, is the answer you got after furthermore the orthogonally diagonalized matrix A? $\endgroup$ – Kenneth Hend May 5 '13 at 22:50
  • $\begingroup$ @Librecoin :What to do if the 2 eigen vectors corresponding to a eigen value aren't orthogonal i.e they do not form a basis i.e the dot product isn't zero? $\endgroup$ – Krishna Deshmukh May 5 at 8:33
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An algorithm is given on the relevant Wikipedia page: http://en.wikipedia.org/wiki/Orthogonal_diagonalization

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