# What set of properties does a matrix have to have in order for all its pairs of distinct eigenvectors $x$ and $y$ to have $x^Ty=0?$ [duplicate]

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I've often seen this property used when dealing with covariance matrices, which have just have every desirable property imaginable (e.g. they are symmetric, contain only real numbers, and are positive semi-definite), but I don't recall every seeing any use of this property outside of this context. Is it just a useful property of covariance matrices? Or is it a result of some of their non-unique properties, and if so, which properties?

## marked as duplicate by Matthew Towers, B. Mehta, Rahul, Community♦Apr 10 '18 at 21:58

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## 1 Answer

You didn't formulate your question correctly (there is no matrix as in the question, simple exercise: let $y$ be a multiple of $x$).

On the other hand, trying to understand what is your real question you might want to consider normal operators: note that there is always a basis formed by orthogonal eigenvectors (so you have an equivalence to the property that you seem to be considering).

• All $1\times 1$ matrices with coefficients in $\mathbb{F}_2$ have the property they originally stated. – user550675 Apr 10 '18 at 20:42
• Never heard of covariance with other fields... :) – John B Apr 10 '18 at 20:43
• That doesn't change that the claim that there is no such matrix is false as written. – user550675 Apr 10 '18 at 20:49
• I suggest that you give your own answer. – John B Apr 10 '18 at 20:52