5
$\begingroup$

If we have two linear transformations denoted by matrices $A, B$ operating on an arbitrary vector $v \in \mathbb R^n$, then how does $Av$ and $Bv$ differ geometrically from $(A-B)v$ ? Does the difference inherit properties from the two linear transformations, or is there no pattern at all?

The reason why I ask pertains to eigen vectors, for they are sent to the null vector when operated upon by $\lambda I - A$ and I am trying to geometrically understand why. If you can explain this second question and not the first one, that would also be fine.

$\endgroup$
1
  • 1
    $\begingroup$ Not sure if you are looking for something deeper than $(A-B)v=Av-Bv$; in other words, $A-B$ maps $v$ to the vector pointing from $Bv$ to $Av$, and this is the zero vector when $Av=Bv$. $\endgroup$
    – angryavian
    Dec 8, 2015 at 8:42

1 Answer 1

3
$\begingroup$

One motivation for the concept of eigenvectors and eigenvalues is the following question: “Does a given linear transformation map some line through the origin onto itself?” If there is such a line for the linear transformation $A$, then for every vector $\mathbf v$ on that line we must have $A\mathbf v=\lambda \mathbf v$, where $\lambda$ is a fixed scalar. So, if $\mathbf v$ is an eigenvector corresponding to the eigenvalue $\lambda$, then $(A-\lambda I)\mathbf v=0$ simply says that $A$ maps $\mathbf v$ to another vector on the line through the origin that contains $\mathbf v$ itself.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .