If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Linear algebra>Unit 2

Lesson 7: Transpose of a matrix

# Visualizations of left nullspace and rowspace

Relationship between left nullspace, rowspace, column space and nullspace. Created by Sal Khan.

## Want to join the conversation?

• Sorry my poor English

I don't understand why does Sal draw left nullspace in codomain and rowspace in domain.

As far as I know, the nullspace belongs to domain because it's the subspace of the domain that makes all vectors 0 in the codomain. And column space spans all the linear combinations of the column vectors in the codomain.

But why does left null space belong to the codomain? apart from in this particular example the left null space and the domain both are 3rd dimension.
• The domain of the transformation "Ax" in this video is R^3, the codomain R^2 (Because A is 2x3).

The domain of "((A)T)z" ("A transpose times z") is R^2, the codomain R^3 (Because (A)T is 3x2).

This means that the domain of the transformation for the left nullspace of A is the codomain of that for the nullspace of A. What follows here is more detail:

We know that the transpose of a matrix product "AB" is equal to ((B)T)(A)T (expained four videos ago), and so the left N(A) is equal to N((A)T), because:

if ((A)T)z = 0, then [((A)T)z]T = (0)T = ((z)T)[((A)T)T] = ((z)T)A.

The equation for the left nullspace in either of the above forms, "((A)T)z = 0" or "((z)T)A = (0)T" has the same result (transforms an R^2 vector into the 0 vector in R^3), and so is a transformation from R^2 (the codomain of A) to R^3 (the domain of A) even though "((z)T)A" has A in it and not (A)T.

I found it helpful to write out all three systems of equations in vector form ("Ax = 0", "((A)T)z = 0", and "((z)T)A = (0)T"), using the matrix A in this video, and using the correct row and column vectors, then find the solutions with just matrix multiplication (not using augmented matrices, which work only for the first two).
• It seems that the dimenison of the column space = number of rows and the dimension of the row space = number of columns. Does that always happen? Where did he prove this?
• This does not always happen. Simple example:
[1 0 0]
[0 1 0]
[0 0 0]

has three rows, but the dimension of the column space is only two. The row space has dimension 2, but there are three columns.

What is however true is the Rank Nullity Theorem, which says the following:
Let A be an mxn matrix. Then dim(Column Space) + dim(Null Space) = n.
Some proofs of this theorem are very long, but you can definitely find one on google somewhere if you want.
• Sal said he would make a proof for how N(A^T) and C(A) are othogonal in the next video. He didn't. Perhaps he has done it in the next playlist, although I haven't got that far yet. Regardless, I thought I'd share my proof of it anyway.

Let the matrix A = [ a1 a2 ... an ], where ai are column vectors.
The column space of A, C(A), is the span of all the vectors ai.
The nullspace of A^T, or the left nullspace of A, is the set of all vectors x such that A^T x = 0. This is hard to write out, but A^T is a bunch of row vectors ai^T.
Performing the matrix-vector multiplication, A^T x = 0 is the same as ai dot x = 0 for all ai.
This means that x is orthogonal to every vector ai.
This means that every member of the left nullspace of A is orthogonal to each of the column vectors of A.
Because the dot product is a linear operation, this means that every member of the left nullspace of A is orthogonal to every linear combination of the column vectors of A, i.e. every member of the span of the column vectors of A.
Therefore, N(A^T) is orthogonal to C(A).
QED.
• I looked at it this way - unless I'm missing something, it's not an oversimplification:

C(A) and the rowspace of A^t ("R(A^t)") are identical. Since N(A^t) is orthogonal to "R(A^t)", it's also orthogonal to C(A). QED.
(1 vote)
• Is this another way of stating what's in the video?
Given Amxn, the preimage of 0 in R^n under A^T is the "orthogonal complement" of the image of A.
• The preimage of 0 under "(A)T (A transpose)" is in R^m.

So "The preimage of the 0 vector in R^m under (A)T is the orthogonal complement of the image of A.

Or, "N((A)T) = the orthogonal complement of the rowspace of (A)T.
• As far as I know, the column space of a matrix is equal to the its row space. Doesn't this equality have implications on the equalities derived in this video? Namely:
1)row space(A) orthog on N(A) <=> C(A^T) orthog on left N(A^T)
2)C(A) orthog on left N(A) <=> row space (A^T) orthog on N(A^T)
(1 vote)
• Well the column space is not equal to the row space. They don't even need to have the same number of elements -- as shown in this example in the video.
(1 vote)
• I realized that the cross product of the two vector that spans N(A) is (1, -1/2, -3/2), which is one vector of C(At), which is the span of (2, -1, -3). And the graph that Sal drew also suggest a relationship to cross product. Is that correct that the cross product of the span of N(A) is always part of C(At)? And what is the deeper meaning of this? Thank you in advance!
(1 vote)
• The cross product always returns a vector which is othogonal to the input vectors. So, if N(A) has dimension 2 and is a member of R3, taking the cross product of two independent vectors within N(A) will give you a vector in C(At), because C(At) is always orthogonal to N(A).

However, this observation only applies if N(A) has dimension 2 and is a member of R3, otherwise it is not possible to take the cross product of two linearly independent vectors in N(A) to obtain a vector outside N(A).
(1 vote)
• Interesting video. A bit convoluted and overextended. Would my final conclusion be correct?

Finding the Nullspace(Transpose(A)) is the same as trying to figure out some set of vectors V that are an orthogonal complement of the row vectors of the original matrix A.
(1 vote)
• in this situation, left space and column space are the lines. However, there exists any situations that both of them are R2 or one of them is R2?
If it does, i think they are not orthogonal, right??

p/s: sorry for my bad grammar
(1 vote)
• The left null space = N((A)T) (the null space of A transpose). C(A) = the row space of (A)T.
As you will see later, the row space and null space of a matrix are always complementary. The domain of (A)T is R^2 so dim(N((A)T)) + dim(Rowspace((A)T)) = 1 + 1 = 2.
If the domain of a matrix is R^4, its row space and null space dimensions must add up to four. So both could be two dimensional. If the domain is 3 dimensional (as in A in this video, where dim(N(A)) is 2, and dim(R(A)) is 1, they must add to three (as this video shows), and n; n.
(1 vote)
• What was he referring to when he said there was a mapping from R3 to R2?

How did he come up with this?
(1 vote)
• What does "complement" mean in terms of "orthogonal complement"?
(1 vote)