linearly independent.

linearly independent.

linearly, let me just write, linearly independent.

So they're linearly independent.

T is linearly independent.

Sorry, not linearly independent.

So it's clearly linearly independent.

Then you are linearly independent.

So they are linearly independent.

So these are definitely linearly independent.

These three columns are clearly linearly independent.

So it has n linearly independent vectors.

linearly independent, they obviously span our column space by definition, and they all our linearly independent, they form the basis.

But we weren't clear whether this was linearly independent, and if it's not linearly independent, it won't be a sufficient basis.

It's pretty obvious from inspection they're linearly independent.

If we are linearly independent, the nullspace of

All pivot columns, by definition are linearly independent.

So all of these guys are linearly independent.

Well let's assume that it isn't linearly independent.

Let's assume that A has n linearly independent eigenvectors.

So my question is are these guys linearly independent?

And we used that information that they're not linearly independent to try to make them linearly independent by getting rid of the redundant ones.

A basis is a spanning set that is linearly independent.

So we also know that we have linearly independent columns.

So it's very clear that these guys are linearly independent.

linearly independent, then that's the 2 constraints for a basis.

These things must span Rn and they must be linearly independent.

So these vectors, and we just said that they're linearly independent.

Now let's say that we have n linearly independent eigenvectors of A.

We're just assuming that A has at least n linearly independent eigenvectors.

It's clear that the set r1, r2, and r4 is linearly independent.

V1 and V2 are linearly independent, that's by definition of a basis.

We have vector V2, which is linearly independent from V1 and U1.

But if this subset of A spans V, then A becomes linearly independent.

linearly independent the only solution here is all of these equal to 0.

You have to span the subspace, and you have to be linearly independent.

So if B is orthonormal, B is also going to be linearly independent.

Now, n linearly independent vectors in Rn can definitely be a basis for Rn.

If we're linearly independent, then the nullspace of A only contains the 0 vector.

That all of the pivot columns in reduced row echelon form are linearly independent.

If A has n linearly independent eigenvectors, and this isn't always the case, but we can figure out that eigenvectors and say, hey, I can take a collection of n of these that are linearly independent, then those will be a basis for Rn. n linearly independent vectors in Rn are a basis for Rn.

Basis means two things it means it spans V and it means it's linearly independent.

And when I say it's linearly independent, I'm just saying the set of pivot columns.

The set of pivot columns for any reduced row echelon form matrix is linearly independent.

Or another way to think about it is that its columns are not linearly independent.