Let's suppose that we have matrix $A_{n\times{n}}$ and we have calculated its distinct (we can additionally assume also non-zero - if needed for a method) eigenvalues $\lambda_1, \lambda_2, ...\lambda_n$. We are also assuming that corresponding unknown eigenvectors are of unit length (in order to have ${v_i}^T{v_j}$ as the cosine of angle between these vectors).


  • Is there a method of calculating absolute value for cosine of angle between any pair of eigenvectors $v_i,v_j$ without explicit calculation these eigenvectors?

1 Answer 1


Because so far (after one week) there is no answer in general case I present here a solution for a very limited case i.e. for dimension $n=2$.
In this case it is easy to check that wanted cosine between eigenvectors (for eigenvalues $\lambda_1, \lambda_2$) can be obtained by taking any non-zero rows of matrices $A-\lambda_1{I} $ and $A-\lambda_2{I} $, normalizing these vectors to unit length and calculating their dot product. We can only speak here about uniquely determined absolute value of cosine as the eigenvectors can be of both directions $+v$ or $-v$.

Formula for this operation:

$\vert(\cos(v_1,v_2)\vert= \vert\dfrac{e^T_i(A-\lambda_1{I} )(A^T-\lambda_2{I})e_j}{\Vert{e^T_i(A-\lambda_1{I}) }\Vert\Vert{(A^T-\lambda_2{I})e_j}\Vert}\vert$
where $i,j$ are indices for any non-zero rows of listed rank one matrices.

Making such calculation we are fulfilling condition that it's no direct calculation of eigenvectors.


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