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# Matrix condition for one-to-one transformation

Showing that the rank of the of an mxn transformation matrix has to be n for the transformation to be one-to-one (injective). Created by Sal Khan.

## Want to join the conversation?

• Have the terms "homogeneous" and "inhomogeneous" been defined somewhere in the Linear Algebra playlist?
• Ax= 0 is a homogeneous system, homogeneous just means that you are looking for solutions when the right side is zero. When the right side is not zero, then you'd call it inhomogeneous and the 'system' looks like Ax=b, where b is a non-zero vector.
• so in short, if matrix A's column vectors are not linearly independent then it's rank/dimension will be less than n. and if the dimension is less than n, the matrix won't be able to span R^n thus, the transformation of A is not one to one...? am i getting this right?
• Everything you said there is right. Even though there are a lot of different new concepts in linear algebra, they are all closely related to one another, so it's really a good idea to be able to tie them together like you did here!
• I felt good about Linear Algebra up until this video.
IF anybody else had trouble understanding this, could you give any advice on how you came to knowing what was going on?
• sal says that A times xp (particular solution) = B. how was this conclusion reached ?
• the problem I have with this statement is that Xp was also defined to be b' which is just a particular vector that satisfies Ax=b' but now its also a solution?
• At Sal substitutes x for x - x_p, why? I fully understand the mechanical reasoning behind the substitution as it allows you to get to 0, what I don't understand is how doing this allows you to show that ANY solution x has the form x = x_p + n where n is a member of the nullspace. Am I wrong in believing that this can be shown just based on the first part (checking the solution) alone? I don't see how this adds any new information.
• Why is b' called "a particular solution for Ax = b"? As far as I can see, b' is just a vector which we get if we do a rref for augmented matrix [A|b}
• When you find the rref you're solving the system. If b not = 0, sometimes b' is the solution for the system, sometimes it's just one solution from an infinite number. I guess that's why Sal uses "x_p" rather than "b'" (I didn't get that till now - thanks).

For example, if the rref of A is the identity matrix, then b' is a list of the values of all the variables that solve the equation, so b' clearly can be a solution to the system. If there are free variables, you get b' plus a span of vectors with the free variables as variable scalars, and in that case there are an infinite number of solutions, including b'. So if there is a solution, and b not = 0, b' is always part of it.

Could b' turn out to be 0 by coincidence?
• What exactly does the subtitle of the video refer to?

"Showing that the rank of the of an mxn transformation matrix has to be an [sic?] for the transformation to be one-to-one (injective)"
• an = n; where n is from "an mxn transformation matrix". See .
• So this means that a transformation T:V->W is bijective (every element in W is reached exactly 1 time) iff rank(A)=dim(V) (prerequisite for injectivity) and rank(A)=dim(W) (prerequisite for surjectivity) so that dim(V)=dim(W) becomes the prerequisite for bijectivity?