00:01
Let f from r2 to r3 be the linear transformation determined by f at 1 ,0 equal 2 ,5 ,1 and f at 0 ,1 equal negative 3 ,1 ,2.
00:15
In part a we want to find f at 9 ,8 in part b we want to find the matrix of the linear transformation f and in part c we discuss about f being injective, surjective or bijective.
00:31
Good so we see here that the linear transformation is given at two particular vectors 1 ,0 and 0 ,1 and with that we can determine completely the linear transformation because these two vectors where we have been given the value of f are or form a basis for r2.
01:00
In fact this is standard basis for r2.
01:05
Vectors 0 ,1 and 1 ,0 are the standard or canonical basis for r2.
01:13
So any vector can be put in a linear combination with unique coefficients of these two standard vectors.
01:22
So to find f at 9 ,8 we see that 9 ,8 can be put the following way is 9 times 1 ,0 plus 8 times 0 ,1 because if you do this vector operation you get this vector here.
01:53
The first product here will be 9 ,0 and the second one here will be 0 ,8.
01:59
9 ,0 plus 0 ,8 is 9 ,8.
02:03
So being f linear we can say that this is equal, let me get rid of these 9's for a moment.
02:12
So this is equal to scalar 9 times f at 1 ,0 plus scalar 8 times f at 0 ,1.
02:21
This is the linearity of f.
02:25
And now this is 9 times f at 1 ,0 is 2 ,5 ,1 plus 8 times f at 0 ,1 is negative 3 ,1 ,2 and then this is 9 times 2 ,5 ,1 is 9 times each component we get 18 ,45 ,9 then plus negative 24 ,8 and 16.
03:11
And when you add these two vectors you get 18 minus 24, the first component, so component 45 plus 8 and the third component is 9 plus 16.
03:32
So you get negative 6, 53 and 25.
03:46
So f at 9 ,8, that is f at vector 9 ,8 is vector negative 6, 53 and 25.
04:06
And that's for a.
04:10
And all that we had to do was to write the vector where we want to calculate the linear transformation applied to.
04:20
We wrote that vector as a linear combination of the two standard vectors 1 ,0 ,0 ,1 where we know the image of each one of them.
04:33
And we know that can be done and in fact in a unique way because vectors 1 ,0 and 0 ,1 are formed a basis of r2, in fact it's the standard basis of r2.
04:49
So this is the answer in part a.
04:51
In part b we need to find the matrix of linear transformation.
04:57
So you can see that we can do that we did in part a for any vector in the domain r2.
05:09
So for any x ,y in r2 we have that f at x ,y is, so we write x ,y as a linear combination of the standard basis, is x times 1 ,0 plus y times 0 ,1.
05:38
If you do x times 1 ,0 you get x ,0 and y times 0 ,1 is 0 ,y.
05:45
So x ,0 plus 0 ,y is x ,y.
05:49
And now we apply linearity of f that is x times f at 1 ,0 plus y times f at 0 ,1.
06:02
And now you replace these two images that have been given.
06:06
So this is x times x times and f at 1 ,0 is vector 2 ,5 ,1 plus y times vector at 0 ,1 is negative 3 ,2.
06:31
Now you do the products here, it's 2x, 5x, x is the first product here, plus the second product here is negative 3y, y and 2y.
06:50
Now you do the sum of vectors and you get 2x minus 3y, first component, 5x plus y, second component, and x plus 2y, the third component.
07:11
And this is evident that this is the matrix 2 ,3, 5 ,1, 1 ,2 times vector x ,y.
07:22
Simply because if you do the product matrix vector here, you get 2x plus 3y, i think i made a mistake, it's negative 3.
07:32
So 2x minus 3y here, 5x plus y here, and x plus 2y here.
07:41
So when we do that, that is we write the image of any vector in the domain of f as a matrix, a constant matrix times vector, that matrix we get is the matrix of the linear transformation.
07:58
So the matrix of the linear transformation f is matrix a equal to negative 3, 5 ,1 and 1 ,2.
08:30
And of course, being the linear transformation known at the standard basis vectors 1 ,0 ,0 ,1, we can say, we could have said directly that the matrix of the linear system, of the linear transformation, sorry, is the matrix whose columns are the images of these two standard basis vectors, that is 2 ,5 ,1 and negative 3 ,1 ,2 as we see here.
08:59
So we have part b already, this is the answer of part b.
09:06
And now in part c, we discuss about the properties of the transformation f, whether it is injective, surjective or bijective.
09:19
Okay, so one thing we get to keep in mind is that if a linear transformation is injective, then the only vector that can go to, that can have image 0, is the 0 vector.
09:41
Of course, the image of the 0 vector is 0, but if it is the only that has that property, the linear transformation is injective...