all of this is kinda goofy when your field isn't algebraically closed, and need not have eigenvalues/vectors.

mad: the algebraic multiplicity of k as an eigenvalue of the nxn A is the geometric multiplicity of 0 as an eigenvalue of (A - kI)^n. this is a 'cheap' way of making algebraic multiplicity 'geometric,' at the cost of changing the operator you're talking about.

you are talking to a wall

i have no idea what you going on about

just drop it lol

i hoped for simple answer that i can understand with my SMOL brain

OK. i assume by default that if someone asks about what the algebraic multiplicity means, they know a little something about what it is. it is better than a default assumption that you don't know anything.

come back to it in a while.

will do for sure

there's no simple answer that i know of. there's something that people do call the "geometric multiplicity" of an eigenvalue, and it generally isn't the algebraic multiplicity.

so the fact that there's something that is (1) not the algebraic multiplicity and (b) is actually called the "geometric" multiplicity, is an informal suggestion that the algebraic multiplicity might not be very geometric.

although everything is eventually geometric, to hear the way people go on about geometry around here.

@leslie Well, if you graph the characteristic polynomial, you can interpret algebraic multiplicity geometrically (order of contact of the graph with the horizontal axis at the root).

Jeeze TEDDY be crushin it

now thats something i can imagine and understand!!!

smol brain satisfied : )

school break

i need to get done all pending intellectual duties asap to get back to math

mad spaces will now 'see' the difference between 4th and 6th order contact on a graph.

9 months mathless, what will that feel like

shin: a neverending party.

@leslietownes dont take it away from :(

i have econ papers i want to read but can't because i've been mathless too long

a mildly frustrating neverending party

Wiggle the polynomial $\varepsilon$?

one dude at my physics department won a prise for having a good disertation

i remember having him as a tutor in my junior year

He didnt know the difference between Sin^2 x and sin x ^2

So i am very shocked

But you know, dirty physicists

@MadSpaces I think Sheldon Axler does it in terms of "generalized eigenvalue"

(if $(T-\lambda I)^k v=0$ for some $k$)

maa.org/sites/default/files/pdf/awards/Axler-Ford-1996.pdf

akiva, we tried this earlier. it didn't work.

Ah arright

What are the eigenvalues/vectors of $\begin{bmatrix}1&1\\0&t\end{bmatrix}$? What happens as $t\to1$?

I guess it's $(1,0)$ and $(1,t-1)$?

You have an extra &.

So I do/did

So now let's write a random vector $(x,y)$ in the basis $(1,0)$ and $(1,t-1)$

It equals $(x-\frac y{t-1})(1,0)+\frac y{t-1}(1,t-1)$

Probably I guess it's most useful to write $(0,1)$ in this basis

$(1,t-1)\frac1{t-1}-(1,0)\frac1{t-1}$

I guess the picture hear is that a shear is the limiting case of two eigenspaces getting closer and closer together

and therefore the algebraic multiplicity counts those two eigenspaces that "merged"

(or you could imagine it as two infinitesimally close eigenspaces)

(the geometric multiplicity counts the two "infinitesimally close" eigenspaces as the same, but the algebraic multiplicity is more discerning)

Next you’ll bring hyperreals into the story!

Let $t=1+\epsilon$ where $\epsilon$ is an infinitesimal hyperreal

Then $(0,1)=(\frac1\epsilon,1)-(\frac1\epsilon,0)$ where those are eigenvectors

@TedShifrin Done^

I was responding to your request

Ultimately, we should understand this for roots of polynomials. A root of algebraic multiplicity $k$ occurs when $k$ nearby roots coalesce.

Interesting versions of this in differential geometry, too.

You know what this reminds me of? The solution space of $y''=0$ is a limiting case of $y''=cy'$ as $c\to0$

because $x=\lim_{c\to0}e^{cx}/c$

Well, look at the $e^x$, $xe^x$ solution set of $y”-2y’+y=0$.

diagonalizable matrices are zariski-dense in all matrices, so it suffices to interpret algebraic multiplicity geometrically for diagonalizable matrices. that case, however, is trivial.

Right - $xe^x=\lim_{b,a\to1}\frac{e^{ax}-e^{bx}}{a-b}$

or something, I didn't check

@Thorgott “suffices”

algebraic geometers would call this an honest argument

I’m an algebraic geometer who does not.

That's basically what I did - found a limiting sequence of diagonalizable matrices, and saw what happens in the limit

I’ll accept that (plus continuity) as a proof of Cayley-Hamilton and various things.

We’ve forgotten the asker of the question.

Yeah, whoops

I did it

I found the BEST theorem

(A directed graph is Eulerian iff it's connected and for every vertex, indegree equals outdegree)

(A spanning tree of a directed graph is "rooted at v" if every edge of the tree points towards v, or in other words, the paths towards v all follow the orientations of the edges)

A corollary is that $c(G,v)$ is independent of $v$

The proof is straightforward - choose v and an edge starting at it called e - it bijects it with a choice of spanning tree rooted at v and an ordering of the remaining out-edges of all the other vertices