Alright, so in this video. I want to talk about what are known as one-to-one and onto functions So eventually I'm going to start talking about this in a linear Algebra setting we'll talk about vectors and linear transformations But at the beginning, you know this is something I'm going to do it a little more basic and even just something you know if you're in Algebra you could understand this you know high school Algebra, so Okay, so a linear transformation Is a transformation t from a vector space V to w and it's called? one-To-one if t maps distinct vectors in V to distinct vectors in W So if you start with a unique vector, you know if you start with two different vectors when you apply that Transformation you get [to] two different outputs as all. It says of [its] [one-to-one] if The range of that transformation equals all of w t is called on [two] So the idea is if you take any vector in in the vector Space w There's some vector in V that it's going to get mapped to [it], so You see this this next little these next little diagrams even in Algebra again So let's talk about this and we'll talk about a couple just basic examples. Hopefully to make this a little more a Little [More] familiar to you So okay [so] the first one we've got our vector space V and our vector space [w] and our transformation t So a one to one function So if I start with you know the dots represent distinct elements in V suppose the first element just gets mapped right over okay? So the first vector gets mapped to a distinct vector The second one gets mapped to a distinct one and the third one also gets mapped to a distinct one this will be one to One and onto because every distinct vector gets mapped to somebody distinct And I've only got three elements in my set w and all of those have a corresponding vector in the domain So that's one to one and onto Let's look at my next example [here], so suppose my first vector gets mapped to the first element here Okay, let's take another one and just map it over here and let's take our third, and just map it down here This is now one. It's still one to one because distinct Elements get mapped to Distinct outputs It's not on two though because you have vectors or you have elements [in] w that Where there's nothing that's getting mapped to them so Not everything in w is sort of getting hit so that means it is one to one in this case because [they're] distinct But it's not on two because you're missing some stuff so in our Last example here suppose my first element gets mapped to the first one Suppose the second element also gets mapped to that first one and then my third Element just goes to that other one This is not one-to-one, [and] [it's] not one-to-one [because] I have these two distinct elements that are going to the same Element the same Vector in W so that's that makes it not one-to-one It is on two though because I only have two elements in w and each of those have a corresponding vector That's getting mapped to them so if you think about this just in terms of of a You know Algebra an example of a one-To-one an onto function? you know suppose we have a function f of x Equals x to the third, so that's going to be my function. [that's] your transformation You're taking an element in [R1] some number x and it's going to get mapped also to R1 So really we're doing to your transformation from R1 to R1 is what we're doing If you want to write it think about it and a linear Algebra setting and I think some places I am it But again all of our transformations here are going to be linear transformations So x to the third recall looks just like that Roughly like that if I start with distinct Elements, so here's some X1. Here's some x2 they get mapped to different elements on the y axis that makes it one to one Okay, [so] if I start [with] this deep distinct elements at the beginning I get distinct y-values out at the end So this would be one to one notice it's also on two because if you take any element Along the y axis you can always find some corresponding element on the x axis that gets mapped to it So that would be one to one and on to Let's look at another one So let's look. I think our next one was it's one to one but not on two suppose our function f of x suppose f of x equals arctangent of x so we start with some number and R1 and we take the inverse tangent of that number well if you recall what arctangent looks like it has an asymptote [a] horizontal asymptote at PI over two and also at negative Pi over two Arc tangent hey looks like that, and it looks like that so this one again is going to be one to one because if I start [with] two distinct elements say [X1] [x2] if I start with Distinct elements again I get if I start with distinct x values I get distinct y values out okay. There's no There's no Two different x values that go to the same y value, so this is [one-to-one] But notice it's not on - it's not on to the you know on the R1 because you're only getting values Greater than negative Pi over [two] and less than Pi over two so we're only getting this little this little interval of values out from negative Pi over 2 to positive Pi over [two], so It's not on [two] Because for example. There's no, there's no element that gets mapped to say up here to positive 10, okay? So in this case. It's not on two, but it is one to one and I think our last one we had a function That's not one-to-one, but is on [two] well again. There's lots of these I think I said let's let f of x equal x to the third minus x So that has x-intercepts at 0 1 and negative 1 so again not quite to scale here To the 3rd minus x looks something roughly like that Okay, so this is not one-to-one. It's not one-to-one Because there are distinct you know so suppose. I look at this [y] value suppose suppose, okay? That's the corresponding y value there's some x value over here that gives me that corresponding y value but notice there's also another [x] value over here that gives me that same [y] value so [I] have distinct values that I start with X1 and X2 but when I Cue them and then subtract that that value x I end up getting the same y value out that means it's not one-to-one But again notice it is on 2 because the domain of this dysfunction it goes from negative infinity all the way to positive infinity So [it] is on 2 so just kind of three examples So this is kind of the basic idea of one-to-one and onto functions. [I] should say if a function is one-to-one And onto it has a special name. [it's] called an isomorphism I'm not really going to talk much about isomorphisms in this video But they turned out to be very important in mathematics the idea of a function is one-to-one and onto The ideas usually it ends up preserving some important structure to the original from the original well I guess I should say it preserves structures from from one vector space to the other and that ends up being important But we'll talk [about] that another day So let's look at two examples now. Let's look at an actual Let's do one where Two examples [we] want to determine if our transformation is one-to-one, and if it's on two okay, so here's example [one] now I should give a little disclaimer here. There's there's other ways to do these as well You can actually start talking about the kernel of the transformation And do some stuff with [that] again We'll look at this some other time But here's just sort of a straightforward algebraic way to determine if something is one-to-one and onto But again, not necessarily the most efficient at all times Okay, so we want to know if the transformation is one-to-one that's first thing we'll start with Okay, so suppose that suppose we start with some element, we'll call it x sub 1 and y sub 1 and I take the transformation of that well according to our formula here It says we would get 2 x sub 1 minus y sub 1 and then we would have x sub 1 plus 2 y sub 1 and Let's suppose. I start with another another Another vector, and I take the transformation of that So the transformation of the vector x 2 y 2 so again, we're going from R2 to R2 here well, that would be 2 x sub 2 minus y sub 2 x sub 2 plus 2 y sub 2 But let's suppose Let's suppose that those So some I'm starting with what I think are distinct vectors, but let's suppose at those two the transformation of those two vectors I End up getting the same output so the question is are these what I started with were they unique at the beginning, so Suppose the outputs are the same what we're going to do is try to determine if this is a one-to-one function So if the transformation of these two vectors is equal that would mean that The outputs are equal so that would mean well that So that would mean [that] 2 x of 1 minus y [sub] 1 and x sub 1 plus 2 y sub 1 that equals well 2 x sub 2 minus y sub 2 and x sub 2 plus 2 y sub 2 So these two vectors the outputs end up being the same So this would be the result well now we can just do some Algebra here So the corresponding entries would have to be the same so I would know that 2 x 1 minus y [1] equals 2 x 2 minus y 2 and likewise I would know that x 1 + 2 y 1 Equals x 2 + 2 y 2 now there may be some more efficient ways to do this But this is the way I did it and it worked so I called it a day, so I'm going to take my first equation here and I'm going to solve this for [y] 1 I'm just basically working with [a] system of equations. All I'm going to do now Okay, well if I solve for y 1 I could add y 1 over to the right? on the left I still have 2 x 1 I could subtract 2 x 2 and then I would add y 2 So I've now taken my first equation and solved for y 1 and Now all I'm [going] to do is just plug all of this stuff in for [y] [1] into my second equation [ok], so if I do that I Would have x 1 plus 2 times y 1 which is what we just solved for so 2 x 1 minus 2 x 2 plus y 2 and that's going to be equal to x 2 [+] 2 y 2 Now at this point. [I] saw this and I said well, let's just see what happens, and I started multiplying it out So I've got x 1 + 2 x 2 I'll have 4 x sub 1 minus 4 x sub 2 plus 2 y sub 2 on the right, I still have x 2 plus 2 y 2 Well, let's see what's going to happen here okay? So notice what I've got here I've got 2 y sub 2 + 2 y sub 2 so I could cancel those out from both sides If I combine my like terms notice on the left I would have 5 x sub 1 now I could add the 4 x sub 2 to both sides and that would give [me] 5 x sub 2 on the right side well if you divide by 5 that means that really X1 equals x2 So it says ah if you've got the same output here the [x-coordinates] really what you started with those those first coordinates those first elements were actually the same thing they weren't distinct at all and If we take this and if we plug you know this into either equation you know take either equation If you replace the fact that x sub 1 and x sub 2 [are] actually the same [well] Then - X sub 1 minus y1 equals well - x sub 1 minus y 2 so now I'm just replacing the fact that x sub 2 is really the same thing as x1 Well again, you can subtract those from both sides and that says negative [y1] and negative [y] 2 are equal which means hey Y1 and y2 are equal, [so] [what] this says is it says if you if you end up with the same output? it says if you if you end up with the same output It means that [that] inputs had to be exactly the same and this means that it is [one-To-one] So if the outputs are the same it means hey you had to start with the same input. That's the only way it can happen Ok, so this is kind of a common little technique to show that things are one-to-one You kind of use different variables you set them equal and then you show at the end of the day. They really had to be identical at the beginning [okay], so the next question is is it on - [ok] so is it on - Okay, so I'm going [to] take some generic vector. I'm going to call [that] a b And that belongs to r - I'm going to figure out is there's some element under this transformation so I want to take the transformation of Some vector x y and I want to be able to get out the vector a be whatever I pick out Okay, well Again under our transformation. It says we have 2x [-] y and then we have x + y that was the the formula for our transformation and that [gives] us a and B and again the same thing. We're just going to make a little system of equations so 2x minus y equals a x Plus y equals b Which one did I solve for it really doesn't matter here? I think actually what I did so you could solve for one of the other I actually just took this and added these together So I've got 3x equals a plus b So [I'm] just doing a system of equations 2x plus x is 3x negative y plus y hey those cancel a plus b Okay is a plus b. So [if] we divide we'll have x equals a plus b over 3 and now if we go [back] now if we go back and substitute that into I use the second equation, so [it] says x is going to be a plus b over over Excuse me. What am I doing [here] so x is going to be a plus b over 3? Plus y that equals b. So I'm going to do the same thing here and solve for y So if you solve for y you can multiply both sides of this equation by 3 that would give us a plus b Plus 3y equals 3 B. So I'm just multiplying again both sides by 3 And if I solve this for [y] the first thing I could do is subtract B from both sides So if I subtract B from both sides I would be left with 2 B and then I could subtract the a over as well, and then if I divide I'll get to the minus a divided by 3 So what this tells me now? It tells me [that] this is actually going to be an on to function and what do these two outputs mean it means that if you wanted to get this vector a You know this vector a [b] out as your solution [it] says what you should start with it says the X-Coordinate should should have the formula a plus b divided by 3 and the y-coordinate should have the formula 2 B minus a over 3 so for example for example suppose Suppose I wanted to take the transformation of some Vector so that at the end of the day you know maybe I get the vector [any] picked one suppose I want to get the vector [1] [1] out [I] Want to get the vector 1 1 out well it says the transformation that will do that It says the first coordinate x it should be a plus b divided by 3 well a plus B. [would] be 2 Divided by 3 and then it says the y-coordinate should be [2b] minus a divided by 3 so in that case 2 times B Would be 2 minus 1 would [be] [1/3] [it] says if you take the transformation of this vector 2/3 And 1/3 with components 2/3 and 1/3 [it] [says] you'll get the vector 1 1 out as your solution [so] it's now giving me a formula. It says you know pick your favorite a pick your favorite b You know and and I can find you a corresponding vector that's going to get mapped to that vector so this now shows that it is on 2 So hey this transformation that we started with is in [fact] a one-to-one and onto function which means again it is an isomorphism Ok so again. There's more efficient ways [to] do this but This is kind of a generic way that the Jews And other settings as well, so let's look at one more example here suppose. I have a transformation from R2 to R3 Where I take a vector with components x y and then I get a vector with components for x x minus y and 0? same thing I want to know if it's if it's [one-to-one] and is it on - well the same thing if I take a transformation So we'll assume. Maybe hey these are these distinct. I start with X1 and y1, and I take some other Some other vector with components X2 and Y2 and I take the transformation of that, but I get the output is the same Well that means that for X1 X1 Minus Y1 + 0 it says that's the for the same thing as 4 x 2 x 2 minus [Y2] 0 and now you can do the same thing So [this] one's a little bit easier [if] you set up your system of equations. It says 4 [x] [1] equals 4 x 2 It says X1 Minus Y1 equals X2 minus Y2 ok so 0 [equals] 0 but from this first one it's easy to see that well if you divide by 4 the x 1 has to equal x 2 and if x 1 and x 2 are the same well then we could [just] subtract them off from both sides and that Means that negative [Y1] equals negative y 2 which it gives that means that y 1 and y 2 are equal, so yes? It is certainly one to one now the question is is it on - Okay, well this one's pretty easy because it can't be on - because If you take a you know some vector in R3 suppose. I examine it. You know suppose you consider So the vector suppose I look at the vector with components 0 0 and pi that's certainly a vector in R3 but by definition This transformation says that no matter what you start with it says this last component By definition is always going to be equal [to] 0 because that's how we set it up So [it's] not going to be on - because for [example] this vector that lives in R3 There's no vector from [R2] under this transformation that's going to get mapped to it so all the vectors under this transformation All have a last component of [zero] which means we can never get Vectors in R3 with where the last component is not zero again, just by construction so this would be one that [is] in fact not on - So okay, that's my my exposé on one-to-one and onto functions. I hope this makes a little bit of sense again You can talk about the kernel and do some other things to consider some of this stuff And we can do some of those but as always I hope this helps feel free to post any comments and questions Hopefully either I can steer you in the correct direction [or] somebody else out there camp
