Computing the Four Fundamental Subspaces

MIT 18.06SC Linear Algebra, Fall 2011
Instructor: Ben Harris

A teaching assistant works through a problem on the four fundamental subspaces.

License: Creative Commons BY-NC-SA

[StanislavBashkirtsev]

He was able to take an inverse of L matrix by changing the sign of the entries. But this only works for atomic lower triangular matrices (those that have 1 column filled and others columns have only diagonal entries).

[des]

1. B changes null space of B(n(B) into zero vector. this is definition of null space.
2. B changes Btranspose into BBtrasnspose . then the column space of BBtranspose would be equal to the column space of B. so you can say the column space of Btranspose changed into column space of B.
3. vice versa for B transpose

[cshsc]

For those of you who are confused about the last picture, null(B) = span{(-3/5, -1, 1)} (not -3/2), and Image(B) = span{(0, 1, 0), (1, 2, -1)} (not 2, 2, -1). Also, (5, 0, 3) and (0, 1, 1) will not be directly mapped to (1, 2, -1) or (0, 1, 0), but they will lie in plane spanned by (1, 2, -1) and (0, 1, 0). Vice versa holds. If you are still confused, think it in this way.

We have null(B) = (-3/5, -1, 1), and there is a plane orthogonal to null(B). This plane can be formed by linear combination of (5, 0, 3) and (0, 1, 1).
Under linear transformation B, the vectors on the plane will be mapped to column space of B (this is what column space is; column space of a matrix is image of a linear transformation). We've verified that left null space is orthogonal to the column space. Same fact under the mapping B^T also holds.

Hope it clarifies your confusion.

[freeeagle]

The recitations are designed in a way that refreshes the concepts in the previous lecture and meanwhile links to the key concept in the next lecture. Beautiful! I can see the next lecture will probably expand the idea of new vector space Professor Strang briefly mentioned at the end of the previous lecture. That may be the reason why Ben wrote the matrix in the form of LU where L and U can be considered as two spaces multiplied. Other TA's did the same thing of connecting lectures. The whole course has been designed like a delicate and complicated machine. Amazing!

[runningcat]

To find N(B^T), there is another way that does not need to get inverse of L. We can write B^T=(U^T)(L^T), then the problem goes to find null space. But be attention, (U^T) is not invertible, its 3rd column is all 0, which means U^T is not a sequence of elimination. Actually the 3rd line of (L^T) can be seen as equivalent to all-0 so that U^T can be seen as eliminations by ignoring the 3rd column. Then we can find rref(L^T) easily and get the final result.

[averagemarcin]

Thanks for the recitation ! Only thing, I didn't understood Ben's explanation on how to find the basis of the left nullspace.

In the book (Introduction to the Linear Algebra by prof. Gilbert Strang), professor provides a clearer explanation:

"EA = R says that the last row of E combines the three rows of A into the zero row of R. So that last row of E is a basis vector
for the left nullspace. If R had two zero rows, then the last two rows of E would be a basis."

[aram]

Just like how when you multiplying the matrix from left to right, you essentially take linear combinations of rows. It works exactly the same way when you multiply from right to left, you take the linear combinations of the columns of L. That's why you can take the 2 columns of L as the basis for C(B)

[josephlevine]

Everyone confused by the picture: check out 3blue1brown's linear algebra series. It takes an approach of viewing matrices as transformations as opposed to thinking about solutions to equations like we have been seeing in class

[sahebsarkar]

8:04 is the last row or basis of left null space will be [-1 0 1] I calculate the B matrix it is [[5 0 3] [10 -1 5] [5 0 3]] . the main thing is when I try to calculate the inverse of L then I get the last row as [-1 0 1] .

[khaledsalah]

10:40 Sir, That Picture was more helpful than the whole lecture, Thanks!

[akmam]

Exciting to be able to look at the same problem from numerous angles! Wow.

[runningcat]

Mr Harris solved the problems in a fast way. Guys, we'd better think it through.

[lee_land_y]

thanks. Can someone explain why pivot columns of L are the basis for C(B). my answer was pivot columns of B

[dawidmachalica]

The video really helped me to systematize the last few lectures. Thank you!

[itsnotthattough]

Thanks a lot for always pointing out precious insights!

[anmol_dot_ninja]

tbh the picture at the end made me more confused

[arunothpoldebnath]

Many of you are complaining about not getting the last picture. Just take a simple example first then come back to 3X3 matrix. Take a 2X2 matrix where the second row is multiple of first. All the four subspaces will be of dimension 1, they will be straight lines. So just draw them on paper with proper measurements and you will see the important things there. Then extrapolate the idea for this 3X3 case.

[alitaylanakyurek]

Why a basis for column space is the columns of upper triangular matrix?

[the_eternal_student]

I think the mapping picture and the inverse were confusing. I understood the lecture well enough, but I did not understand the mapping and the inverse was not arrived at by row operations of the matrix E or any apparent operation.

[henryzhu]

Confused about the last diagram. How a column space is transformed to row space by B?