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How do you orthogonally diagonalize the matrix A?

Matrix A =

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

Kenneth Hend
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    Did you know that eigenvectors of distinct eigenvalues are mutually orthogonal? – Easy May 04 '13 at 03:13
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    An algorithm is given on the Wikipedia page: http://en.wikipedia.org/wiki/Orthogonal_diagonalization – Potato May 04 '13 at 03:14
  • SEE http://math.stackexchange.com/questions/375711/eigenvector-of-matrix-of-equal-numbers/375725#375725 – Will Jagy May 04 '13 at 03:38
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    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. – Gerry Myerson May 04 '13 at 03:54

2 Answers2

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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,

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

Librecoin
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  • 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? – Kenneth Hend May 05 '13 at 22:50
  • @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? – Krishna Deshmukh May 05 '19 at 08:33
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An algorithm is given on the relevant Wikipedia page: http://en.wikipedia.org/wiki/Orthogonal_diagonalization

Potato
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    in case of interest; there is a pattern for these that can be used for any size... http://math.stackexchange.com/questions/375711/eigenvector-of-matrix-of-equal-numbers/375725#375725 – Will Jagy May 04 '13 at 03:56