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I was looking at an interesting visualization of eigenvalues and eigenvectors. The picture is below. I understand the meaning and computation of eigenvalues and eigenvectors, but I think I learned about them in a much more algebra forward way instead of a more graphical way. Hence my question.

enter image description here

Say I have a square 3x3 matrix $A$ that is made up of component vectors $a_1, a_2, a_3$. And this matrix has eigenvectors $v_1, v_2, v_3$ that correspond to 3 eigenvalues. I was wondering what the mathematical relationship was between the directions of $a_i$ and the corresponding eigenvectors of the matrix $A$. From the image it seems like the eigenvector--at least in this case, is a bisector between say $a_1, a_2$.

I am not sure if that holds in all or most cases. But seems like there should be something here related to the dot product between the $a_i$ vectors--which gives the angle between the vectors, and then the corresponding angle between each $a_i$ and the eigenvectors.

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I presume you mean $a_i$ are the columns of $A$, i.e. $a_i = A e_i$ where $e_1, \ldots, e_3$ are the standard unit vectors. If $v_i$ are the eigenvectors, with three distinct eigenvalues $\lambda_i$, and $u_i$ are the left eigenvectors (the transposes of the eigenvectors of the transpose $A^T$) so that $u_i v_i = 1$ for $i=1\ldots 3$, then $$ A = \sum_i \lambda_i v_i u_i$$ so that $$a_j = A e_j = \sum_i \lambda_i v_i u_i e_j $$ That is, $a_j$ is a linear combination of the $v_i$, where the coefficient of $v_i$ is $\lambda_i$ times the $j$'th entry of $u_i$.

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  • $\begingroup$ yes, I see now. I see what you mean. This is the $A = S \Lambda S^{-1}$ decomposition. So the columns of A are expressed as a sum of the left, right eigenvectors as a basis. That is exactly what I was looking for. Thanks for reminding me of that identity. Thanks for your help. $\endgroup$ – krishnab May 4 at 12:59

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