Is this a valid subspace?

So we already know, let's say I have some subspace V. Let's say V is a subspace of Rn.

So my question is, is this a subspace?

Let's say I've got some subspace of Rn.

That's what being a member of the subspace means.

And all bases have the same number of vectors for any given subspace.

Let's say V is a subspace of Rn.

So let's say that B is the basis for some subspace, v.

Let's say that I have a one dimensional subspace. Let's call that V1, for it's a one dimensional subspace.

So just like that, we get a pretty simple way of figuring out the projection onto a subspace with an orthonormal basis.

So this wasn't a subspace, this was just a subset of R2.

And that makes sense, because we said that a column space of a matrix has to be a valid subspace, and a valid subspace has to contain the zero vector.

But it's pretty easy to show that the span of any set of vectors is a legitimate subspace.

So a subspace let's say that I have some subspace oh, let me just call it some subspace s.

In the last video, I spoke somewhat theoretically about what a null space is and we showed that it is a valid subspace.

You're going to have k terms, because this is a k dimensional subspace.

So you can see we have k vectors, so v is a k dimensional subspace.

This was our definition of, or this is our way of finding our transformation matrix for the projection of x onto some subspace.

So let's say I have the subspace V that is spanned by the vectors let's say we're dealing in R4, so the first vector is 0, 0, 1, 1.

But no, your null space will always, by definition, by the fact that it is a subspace, it will always contain a 0 vector.

If we add two vectors in Rm to each other, we get another vector in Rm because Rm is a valid subspace.

So I want to know whether the image of V under T is a subspace.

And in the last video I took great pains to show that all bases for any given subspace will have the same number of elements.

And let me just call my plane, this plane, let me call it, it's going to be some subspace V in R3, which is equal to the span of v1 and v2.

And then finally, to make sure it's a valid subspace and this actually doesn't apply just to column space, so this applies to any span.

So if I have this set, this orthonormal set right here, it's also a set of linearly independent vectors, so it can be a basis for a subspace.

If I take all of the x's if I take the world, the universe, the set of all of the x's that satisfy this equation, do I have a valid subspace?

Proportional

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Partial

This is a subspace if the following are true and this is all a review that the 0 vector I'll just do it like that the 0 vector, is a member of s.

And then a third vector so it's a three dimensional subspace of R4 it's 1, 1, 0, 0, just like that, three dimensional subspace of R4.

We already know that the projection onto any subspace V of x is equal to A times A transpose A inverse, times A, times x.

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And beyond it, space.

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Oh? Part time?

Part time job?

RD

It's only parttime.

So in order for it to be a subspace, if I take the transformation let me find two members of T. Well clearly if I take the transformation of any members of V, I'm getting members of the image.

So we come on the result that T the image of V under T, is a subspace. Which is a useful result which we will be able to use later on.

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