Matrix condition for one-to-one trans | Matrix transformations | Linear Algebra | Khan Academy

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Showing that the rank of the of an mxn transformation matrix has to be an for the transformation to be one-to-one (injective)
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[@beegdigit9811]

Khan always has that eye opening simplicity

[@kasra.b3316]

Very nice video but you could have mentioned that to be "one to one", the matrix needs to be linearly independent at the beginning of the video so that those who know what that means don't have to sit through 17mins of lead up.

[@dhruvrandi007]

No. It's doesn't need to be. Only if it's one one. The result for solution to be sum of particular vector and homogeneous solution is a general one in case where the solution does exist.

[@aidawall8]

your handwriting is beautiful i love it!!! so aesthetically pleasing :) oh, also the video helped a lot. THANKS!

[@ryanlafferty5948]

onto=surjective, one to one = injective, both = bijective

[@blakekl]

What program do you use to draw for these videos?

[@foolio002]

is this about injective and surjective transformations

[@Baradur10]

Whoa, I'm surprised the CC functionality of youtube is working so well.

[@JoranBeasley]

good tutorial thanks man :)

[@NGBigfield]

These videos are so good! and: "let's do this in white!".... "That's not white!!"

[@ReusableDuck]

Why can't my professor teach like this.......

[@nicholasgigliotti3012]

My book says that a matrix can be one to one if it has a unique solution or no solution however, it has says T is one to one if and only if Ax=b has trivial solutions. How can something that only has trivial solutions have no solution. I guess what I am saying is can somebody provide me with a matrix that is one-to-one with no solution?

[@rfmvoers]

"How can something that only has trivial solutions have no solution." You are thinking of a square matrix here. But if m > n, you can have a homogeneous system Ax=0 with only the trivial solution, and an inhomogeneous system Ax=b that has no solutions. Example: x+y = 0, 2x+y = 0, x+2y = 0. This homogeneous system has only the trivial solution (x=0,y=0). Now an inhomogeneous system with the same matrix: x+y = 2, 2x+y = 2, x+2y = 2. This has no solution.

[@jgs474]

noooooooooo concheee en español tbm

[@nenadilic9486]

3:40 Mistake. b' is in Rm and x particular must be in Rn. They clearly are not the same. Another, smaller, mistake at 15:00 - the arrow in the implication has to point the opposite way: if N(A) is NOT only the zero-vector then T is not 1-to-1, and that is what he had proven. The way the implication was shown simply stems from replacing xn with the zero-vector which leaves only one x (xp) mapping to Ax.