Quick question.
How many eigenvectors are there?
If I had more than D images.
Okay.
So D is 10,000, 10,000 dimension,
let's say, or a million.
So if I had, you know, 10,000 images and
I was looking at all those,
I could get 10,000 eigenvectors.
But clearly the sum of
those outer products.
If I've only got a small number, right?
The whole idea is that M is
going to be a lot less than D.
So how many of them are there?
Well your intuition might say,
well I've got M independent things,
so there must be M of them.
But actually your intuition was wrong,
and in fact, a better intuition
would be the following.
Suppose I gave you two points
in three dimensions, okay?
How many vectors can span those?
Just one, right?
Wo, I go between them.
And the reason is, I have to think
about subtracting off the mean.
Right?
And once I subtract off the mean,
actually I only have one
degree of freedom, right?
Because the origin is going to be
halfway between those two points.
So once you tell me where
one of the points is,
I know where the other one is.
So there's only one degree of freedom.
So with two points,
there's only one possible eigenvector.
Let's think about it this way.
If I have three points in space, okay?
How many vectors does it
take to span those points?
Well, they lie in a plane, don't they?
Of course they do.
Three points to find the triangle.
So with three points,
there's only two eigenvectors.
In general,
there are M minus 1 eigenvectors.
And the reason there's M minus 1 is,
I take my endpoints and
I subtract out the mean.
So that removes a degree of freedom.
So that's a trick question that I
always give on the final of my course.
So if anybody's taking the final,
now you know the answer.
Okay, there's M minus 1.
