# Is there a connection between the direction of the vectors in matrix $A$ and the corresponding eigenvectors?

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.

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.

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 $$u_i$$.
• 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. – krishnab May 4 at 12:59