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# Visualizing a column space as a plane in R3

Determining the planar equation for a column space in R3. Created by Sal Khan.

## Want to join the conversation?

• Sorry, I don't have a particular spot in the video to refer to, but here goes:
I have been watching the Linear Algebra sessions from the beginning and thought I had a good grip on what was going on then, when I got here and we had been working with columns for a few lectures, the question came to mind:
How does what we are doing in column spaces relate to the origin of the original matrix. I mean the rows of the matrix represent a linear equation, but the column of a matrix only references the same variable in each of the linear equations.
Hoping for an explanation
Mark
• If you have Ax = b, then b must be a linear combination of the columns of A. Say it is 1 times the first column, plus 2 times the second column... then x is the vector (1,2). For, say, a 2x2 matrix A there is a row way of looking at it, and the lines represented by each row intersect at x, plus a column way of looking at it where each column is a vector and the numbers in x tell you how much to multiply each column by so that their sum results in the b vector. It is 2 ways to do the same problem or think about it.
• I don't get it. Is it always a plane? The column space is only a plane if there are 2 basis vectors. What if there are 3?
• Yes I had the same thought. My guess: if there are three basis vectors, the column space is a volume in R3. If there is one, the column space is a line in R3. As there ar two in this example, it is a plane in R3. Would like confirmation of this, though?
• For the second approach, what about the constraints in the top 2 rows?

e.g., that x_2 - x_3 - x_4 = 2x - y.

Is it not possible that this would alter the plane constraint?

We already heuristically know that the solution should be a plane. Is this why these cases were not (or did not have to be) considered?
• I was wondering the same thing. The last row of the rref of Ax=u (x and u both variable vectors), which is one of the results of solving Ax=u, says unambiguously that 5x - y - z = 0; so do we need more than that?
(1 vote)
• So, since spanning subspaces must always contain the zero vector, then all spanning subspaces intersect the origin?
• All valid subspaces contain the zero vector and therefore pass through the origin.
(1 vote)
• @ sal ,cross products the two positions vectors and those vectors are not "ON" the plane they are just touching the plane . so how can the cross product be a normal to the plane ?

for those having the same question
at 13.34 , Sal clearly explains why he was able to get the normal vector by doing a cross product. ie the two vectors are also part of the plane. now why are they part of the plane because all the points of the vectors can be obtained from their linear combination. eg the equation av1 + bv2 where v1,v2 are vectors and a,b are scalars a can be zeroed out to get the end point of vector v2 and then b can be incremented from zero to get all the points in vector v2. the same thing can be done for v1
• since, origin is lying in the plane so the position vector is also lying in the plane
• Since the rref of the A only has two identity vectors, e1 and e2, doesn't the mean that the column space spans R^2 and not R^3? meaning the plane would be R^2? This kinda goes for the whole video
• R^2 is a description used for the set of all vectors with 2 components, and R^3 is the set of all vectors with 3 components. As these vectors have 3 components they are members of the R^3 set. The column space might then be visualised as a 2d plane inside this set, but it is not R^2 as the vectors still have the extra component.
• At you said the plane has to intersect (0,0,0) because Subspace always contain the zero vector. But i thinks it a little mistake right? Because you said vector doesn't contain position information? Is the real reason is you draw the vectors at the standard point? I also think any other parallel plane with the one that you found can contain the same set of vector.
• I have similar questions. I suspect, but don't expect, the answer will become clear later.
(1 vote)
• At , it is said that the normal vector is the result of the cross product of the two basis vectors. Just to confirm: Is possible then to generalize this by saying that the cross product of any two vectors that are a "basis" (therefore lie on the plane they describe) generate a normal vector? Thus, this wouldn't work with "position vectors" (which do not lie on the plane), right? Finally, is this the main difference between "position vectors" and "basis vectors"?
• Basis vectors are simply position vectors that lie in the subspace, in this case the plane, which are the minimum vectors required to define that subspace. The cross product will find the vector that is orthogonal to that subspace, but you aren't necessarily limited to the cross product of the basis vectors since the other vectors also lie in the subspace. They're simply superfluous vectors in that subspace that can be defined using linear combinations of the basis vectors. You will still end up with the same plane.

Here's proof if you don't believe me (v1 x v3): http://i.imgur.com/S5IvArH.png

Of course, if you just take any old position vector that points to a point on the plane it's not going to work since it doesn't lie along the subspace plane.