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Say there is a square symmetric matrix $\mathbf{M}$ and vector $\mathbf{v}$. Then the vector $\mathbf{\tilde{v}}$ satisfying

$$\mathbf{\tilde{v}' Mv}=0$$

is said to be $\mathbf{M}$-orthogonal to $\mathbf{v}$ (for eg. page 48 of "Portfolio Theory" by Giorgio P. Szego, 1980).

Noting that the eigenvectors of $\mathbf{M}$ also satisfy this relation, i.e.:

$$\mathbf{e}_i' \mathbf{Me}_{j\neq i}=0$$

May I conclude that $\mathbf{v}$ and $\mathbf{\tilde{v}}$ are eigenvectors of $\mathbf{M}$?

Put another way, do there exist vectors $\mathbf{v}$ and $\mathbf{\tilde{v}}$ satisfying the first displayed equation which are not eigenvectors of $\mathbf{M}$?

ben
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1 Answers1

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Let $M$ be a rank 1 projection matrix, say $M=vv^t$ for some unit length $v\in\mathbb{R}^n,n>1$. So $M$ has one eigenvector, $v$. Let $\tilde{v}$ be a vector orthogonal to $v$ and then $\tilde{v}'Mv=\tilde{v}'v=0,$ although $\tilde{v}$ isn't an eigenvector.

snarfblaat
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  • What about symmetric matrices? – ben Oct 17 '16 at 00:49
  • $(vv^t)^t=(v^t)^tv^t=vv^t$ – snarfblaat Oct 17 '16 at 00:50
  • Oops. What I meant to say was: what about symmetric matrices with full rank? – ben Oct 17 '16 at 00:54
  • Let $M$ be square symmetric and full rank $n$ and $v_1,\ldots,v_n$ eigenvectors. So $Mv_1=\lambda v_1,\lambda\in\mathbb{R},$ and you can take a nontrivial linear combination of the other eigenvectors as $\tilde{v}$. The eigenvectors are orthogonal so $\tilde{v}^t(\lambda v)=0$. But clearly not any linear combination is an eigenvector for most $M$. – snarfblaat Oct 17 '16 at 01:03
  • So $\tilde{v}$ must be either one of the other eigenvectors, or a linear combination thereof? – ben Oct 17 '16 at 01:08
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    Yeah but this isn't a real restriction since from our assumption that the matrix is full rank and square, every vector in the space is a linear combination of the eigenvectors. – snarfblaat Oct 17 '16 at 01:10