Geometric understanding of subtracting lambda from diagonals

Question

Given the definition of eigenvalues/eigenvectors:

$Av = \lambda v $

you could rearrange the terms to be:

$(A - \lambda I)v = 0$

Geometrically, the first equation says that multiplying by $A$ is the same as scaling the vector $v$ by $\lambda$. However, in the second equation, how do you visualize the effect of subtracting the matrix $\lambda I$ from matrix $A$ and how does that induce a linearly dependent set of basis vectors?

TL;DR: I understand that the new matrix $(A-\lambda I)$ collapses the span of $v$ into a lower dimension but I don't understand how $A$ relates to $(A-\lambda I)$ geometrically.

Answer 1

May I ask what do you understand by geometric relation or visualization? What you said

$(A-\lambda I)$ collapses the span of $v$ into a lower dimension

is exactly how I visualize it. To be more precisely, it completely removes the image of ${\rm span}(v)$. Thus only the co-space of ${\rm span}(v)$ is possibly nontrivially acted upon.

The action on the remaining co-space is also modified, namely by subtracting $\lambda$ times the input vector from the image, $Aw - \lambda w$ (for $w \perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.