TITLE: $(A-λI)X=0$ works, but$ (λI-A)X=0$ does not when finding eigenvectors. Why?
QUESTION [6 upvotes]: Either works when trying to find the eigenvalues, but only the former works when trying to find corresponding eigenvectors. I can understand how it makes a difference, but what I don't understand how one is supposed to "know" the former is the "correct" form, since it starts from here:
$AX = λX$
And from there, you can end up with either:
$AX - λX = 0$
or 
$λX - AX = 0$
And finally:
$(A-λI)X = 0$
or
$(λI-A)X = 0$
And so should they not both be correct? Furthermore, once it's in that form, can you not multiply both sides by $-1$ to flip them?
I guess I'm overlooking some sort of rule of algebra when dealing with matrices. Let me know. Thanks.
EDIT
Here is an example:
$\pmatrix{5&3\\6&2}$ has an eigenvalue $-4$
If we do $(A-λI)X = 0$:
$\pmatrix{9&3\\6&2}*X = 0$
This solves to $X = \pmatrix{-1\\3}$ (and multiples of it)
If we do $(λI-A)X = 0$:
$\pmatrix{-9&3\\6&-2}*X = 0$
This solves to $X = \pmatrix{1\\3}$ (and multiples of it)

REPLY [2 votes]: You usally use $(\lambda I-A)X$ to calculate the Eigenvalues because the polynom you will get is normalized in this case ( its highest degree coefficient is 1).
For Eigenvektors it totally doesn't matter which way you do it as you have
$(\lambda I-\vec A)\vec v=\vec0 \Leftrightarrow -(\lambda I-A)\vec v=\vec 0 \Leftrightarrow (A-\lambda I)\vec v=\vec 0$