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## Linear algebra

### Course: Linear algebra>Unit 3

Lesson 1: Orthogonal complements

# Rowspace solution to Ax = b example

Visualizing the rowspace solution to Ax=b. Created by Sal Khan.

## Want to join the conversation?

• I realize that I still have some difficulties in understanding the meaning of position vectors.

Starting at Sal explains that the vector [3 0] - r ("the pink vector") is a member of N(A) although it is lying on the solution set of Ax = b ("that pink vector is not in standard position but it is going to be a member in our null space).

Does this mean that every vector is part of both the solution set and the null space? And would this mean that the solution set and the null space are equal? • The pink vector is still part of the null space because shifting a vector from standard position does not change it at all. You could also think of it like this: The solution set to Ax=0 is contained as a subsection to the solution set of Ax=b. I hope this cleared things up for you.
• At , Sal defines r as a member of the column space. Isn't that a mistake - did he not mean that r is a member of the row space? • Watching Grants videos, is the answer for Ax = b the inverse transformation of b?
Ax = b,
A^-1*Ax = A^-1b,
x = A^-1b <- the original vector before making the plane 1d [if not clear ask please :) ]
IF this is the case, how do you get the inverse of this matrix, since determinant is zero?
If this is not the case, were's my falacy? • The null space is the solution set to Ax=0 and in this case it was [2,3]. The solution set to Ax=b in this case was [2,3] + [3,0]. However, why can't the zero vector be b? • The zero vector can be "b" in fact the solution to Ax=0 is called the homogeneous solution; however, the solution to the equation Ax=b is called the general solution and it consists of a linear combination of the homogeneous solution and a particular solution( in this case the particular solution was [3,0]). I suppose the simplest way to put it is: Ax=b is a more general case than Ax=0. I hope this helped.
• Does the formula that Sal relates about 16 mins into the video (i.e. that the basis vector of the column space minus the basis vector for the row space is equal to a vector in the null space) imply the reverse? That we can obtain the vector he calls r ([3 -2]) by subtracting the basis vector of N(A) from the basis of C(A) (i.e. [3 0] - [2 3] = cr)? • If b is a member of C(A) then the shortest solution to Ax=b is a unique member of the rowspace of A. This is a conclusion from a previous video and Sal visually shows it in this video. My question is:

Does saying "If b is a member of C(A)" is equivalent to say "If the equation Ax=b has at least one solution"?

If b is a member of C(A) then we can express it as a linear combination of column vectors b=x1a1+x2a2=Ax where a1,a2 are column vectors of A and x1,x2 are solutions to Ax=b. • So easy after I watched this NOT • Sal used to mention that the solution set of Ax=b is always x=b'+N(A) . Is that suppose to mean that the solution set is always parallel to the N(A)? If that is the case, since the rowspace is orthogonal to N(A), then we can claim that the rowspace is also orthogonal to the solution set, which it implies that the rowspace contain the shortest solution vector.

Video:   