linearly, let me just write, linearly independent.

linearly dependent.

linearly independent.

linearly independent.

So they're linearly independent.

We often think linearly.

A are linearly dependent.

T is linearly independent.

Sorry, not linearly independent.

So it's clearly linearly independent.

Then you are linearly independent.

So they are linearly independent.

But this is definitely linearly dependent.

So these are definitely linearly independent.

These three columns are clearly linearly independent.

Our left hemisphere thinks linearly and methodically.

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.

Now, let's assume that they are linearly dependent.

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.

So we know that this right here is linearly dependent.

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.

These guys are also linearly independant, which I haven't proven.

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

So let's assume that vi and vj are linearly dependent.

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

If I take 30 steps linearly one, two, three, four, five

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

So that implies that you're linearly dependent, which is a contradiction with our original statement that set A is a basis for V, because that means it's linearly independent.

And I'll say this is linearly dependent because this guy spans V.

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.