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Suppose an $n\times n$ matrix has real entries and has $n$ real eigenvalues and its eigenvectors span $\mathbb R^n.$ Are there any interesting conditions under which $k$ of its eigenvectors are mutually orthogonal, for $k<n$? If so, might that conclusion be useful?

PS: Here's an example. This matrix is not symmetric but would become so if the $(1,2)$ entry were changed. Six of its seven eigenvectors are all orthogonal to each other. $$ \left[ \begin{array}{rrrrrrrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 \end{array} \right] $$

Suppose an $n\times n$ matrix has real entries and has $n$ real eigenvalues and its eigenvectors span $\mathbb R^n.$ Are there any interesting conditions under which $k$ of its eigenvectors are mutually orthogonal, for $k<n$? If so, might that conclusion be useful?

Suppose an $n\times n$ matrix has real entries and has $n$ real eigenvalues and its eigenvectors span $\mathbb R^n.$ Are there any interesting conditions under which $k$ of its eigenvectors are mutually orthogonal, for $k<n$? If so, might that conclusion be useful?

PS: Here's an example. This matrix is not symmetric but would become so if the $(1,2)$ entry were changed. Six of its seven eigenvectors are all orthogonal to each other. $$ \left[ \begin{array}{rrrrrrrrrrrrrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 \end{array} \right] $$

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Michael Hardy
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Some but not all eigenvectors mutually orthogonal

Suppose an $n\times n$ matrix has real entries and has $n$ real eigenvalues and its eigenvectors span $\mathbb R^n.$ Are there any interesting conditions under which $k$ of its eigenvectors are mutually orthogonal, for $k<n$? If so, might that conclusion be useful?