Understanding an eigenspace visually. I'm trying to make sense of an example in my textbook but I am confused as to what they are presenting to me. I can do the computations but I would like to have a greater understanding of what is actually going on.
My assumption is that the blue vectors are my eigenvectors but when I cross these two vectors it does not produce the same results. The vectors, when multiplied by A lie off of the plane. My next assumption is that the plane is the $Span(\begin{bmatrix}\frac{1}{2}\\1\\0\end{bmatrix},\begin{bmatrix}-3\\0\\1\end{bmatrix})$. This does not appear to work either.
Would anyone have any insight as to what is going on in this example?

 A: A linear operator is a transformation of the underlying space. A square matrix is a linear operator. Geometrically, it is a rotation, reflection, or inhomogeneous stretching of the space, or combinations thereof. Eigenvalues are conserved directions - e.g. The axis of a rotation or the principal directions of a stretch. If one subtracts a smoothly increasing linear multiple of the identity operator - which is an isotropic stretch of unit magnitude and can be viewed as a unit sphere - from zero, one may imagine the origin "gobbling up" eigenvectors isotropic ally,  almost like a linear black hole! Each eigenvector head is coincident with the origin for precisely one value of the parameter, and this value is the corresponding eigenvalue.  If two eigenvectors have the same length (they cannot be collinear), then they therefore correspond to the same eigenvalue. Hence, any linear combination of them is still an eigenvector - so in this special case there is a whole plane of eigenvectors, in contrast to the "more usual" case of distinct eigenvectors corresponding to distinct eigenvalues. This is termed degeneracy and is manifested, for example, in circular polarisation of light or vibration of surfaces of revolution. It often corresponds to a rotational symmetry or complex structure.  The correspondence between the geometric description and the algebraic one generalises to higher dimensions, mutatis mutandis.Hope that helps :)
A: The plane shown in figure 3 is the eigenspace corresponding the $A$'s eigenvalue $\lambda = 2$. And, as shown in the calculation, $\Bigg\{\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\Bigg\} = \{\mathbf{v_1}, \mathbf{v_2}\}$ is the basis for that space. Any linear combinations of $\mathbf{v_1}$ and $\mathbf{v_2}$ is an eigenvector in that eigenspace. And, the matrix $A$ acts as a dilation to any vectors in that space, i.e. $A\mathbf{x} = 2 \mathbf{x}$ for any $\mathbf{x}$ in that eigenspace. For example,
$$A \mathbf{v_1} = \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} = 2 \mathbf{v_1}$$
$$A (2 \mathbf{v_1} + 3 \mathbf{v_2}) = \begin{bmatrix} -14 \\ 8 \\ 6 \end{bmatrix} = 2 (2 \mathbf{v_1} + 3 \mathbf{v_2})$$
Hope this helps in giving you a visual image of what's going on.
