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Consider the symmetric matrix \begin{equation*} A=\begin{bmatrix} 4 & 2 & 0 & 0\\ 2 & 4 & 0 & 0\\ 0 & 0 & 4 & 2\\ 0 & 0 & 2 & 4 \end{bmatrix}. \end{equation*} I need to find orthonormal eigenvectors of $A$. I have calculated the eigenvalues as $6$ and $2$. For eigenvalue$=6$, I found that one eigenvector is $\begin{bmatrix}1 & 1 & -1 & -1 &\end{bmatrix}^T$. If I perform the Gram-Schmidt process to this vector as well as another vector, say $\begin{bmatrix}-1 & -1 & 1 & 1\end{bmatrix}^T$ I still get to linearly dependent vectors. I am not quite sure what I am doing wrong.

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    $\begingroup$ The eigenvectors (shown as column vectors) for the eigenvalues $6, 6, 2, 2$ should be $$ \begin{pmatrix} 0 & 1 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 1 & 0 & 1 & 0 \\ \end{pmatrix}$$ Now convert them to orthonormal form. $\endgroup$ – Moo Apr 12 '17 at 1:34
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    $\begingroup$ Gram-Schmidt process requires starting from linear independent vectors sets, and results in orthogonal vectors sets. However the two vectors you provided are linearly dependent. $\endgroup$ – user1950580 Apr 12 '17 at 2:46
  • $\begingroup$ You’re starting with a pair of linearly dependent vectors, so it should come as no surprise that the G-S process gives you a linearly dependent result (you should get a zero vector at some point, in fact). $\endgroup$ – amd Apr 12 '17 at 2:59

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