Math - Linear Transformation Question
Yes, this is a LT.
Here is my solution: https://s15.postimg.io/4t7xy4t6j/IMG_1088.jpg
We all have different ways to prove mathematical equations, this is just my own way, if you don't like the traditional way and prefer to prove via matrices, solve it yourself.
Also you will have to excuse my bad handwriting.
Here is my solution: https://s15.postimg.io/4t7xy4t6j/IMG_1088.jpg
We all have different ways to prove mathematical equations, this is just my own way, if you don't like the traditional way and prefer to prove via matrices, solve it yourself.
Also you will have to excuse my bad handwriting.
That seems pretty long.
A function f is a LT iff both these conditions are met:
* f(x + y) = f(x) + f(y)
* f(ax) = a f(x)
T(x, y) = (k1 x, k2 y)
T((x1, x2) + (y1, y2)) = T((x1 + y1, x2 + y2)) = (k1 (x1 + y1), k2 (x2 + y2)) =
= (k1 x1 + k1 y1, k2 x2 + k2 y2) = (k1 x1, k2 x2) + (k1 y1, k2 y2) = T(x1, x2) + T(y1, y2)
T(a(x, y)) = T((ax, ay)) = (a k1 x, a k2 y) = a(k1 x, k2 y) = a T(x, y)
QED
A function f is a LT iff both these conditions are met:
* f(x + y) = f(x) + f(y)
* f(ax) = a f(x)
T(x, y) = (k1 x, k2 y)
T((x1, x2) + (y1, y2)) = T((x1 + y1, x2 + y2)) = (k1 (x1 + y1), k2 (x2 + y2)) =
= (k1 x1 + k1 y1, k2 x2 + k2 y2) = (k1 x1, k2 x2) + (k1 y1, k2 y2) = T(x1, x2) + T(y1, y2)
T(a(x, y)) = T((ax, ay)) = (a k1 x, a k2 y) = a(k1 x, k2 y) = a T(x, y)
QED
It's really not that bad. It may look long but it really took me less than a minute to get the result. As I have stated before, we all have different methods toward proofs. This is just my way, and I think it's a pretty good and safe way.
Looking at it more carefully, you also didn't prove what was asked. All you've shown is that
* T((1, 2) + (3, 4)) = T(1, 2) + T(3, 4)
* T(5 (1, 2)) = 5 T(1, 2)
But this is not what the definition of linear transformation says.
* T((1, 2) + (3, 4)) = T(1, 2) + T(3, 4)
* T(5 (1, 2)) = 5 T(1, 2)
But this is not what the definition of linear transformation says.
You can work with values, you can work with arbitrary variables, all the same.
let v1 = a1, a2
let v2 = b1, b2
vector u + vector v = a1 + b1, a2 + b2
T(U + V) = a1 + b1 / 5, 3(a2 + b2) / 5
= a1 + b1 / 5, 3a2 + 3b2 / 5
T(U) = a1 / 5, 3a2 / 5
T(V) = b1 / 5, 3b2 / 5
T(U) + T(V) = a1 + b1 / 5, 3a2 + 3b2 / 5
It's the same thing, I just like to work with real numbers as oppose to some random variables.
let v1 = a1, a2
let v2 = b1, b2
vector u + vector v = a1 + b1, a2 + b2
T(U + V) = a1 + b1 / 5, 3(a2 + b2) / 5
= a1 + b1 / 5, 3a2 + 3b2 / 5
T(U) = a1 / 5, 3a2 / 5
T(V) = b1 / 5, 3b2 / 5
T(U) + T(V) = a1 + b1 / 5, 3a2 + 3b2 / 5
It's the same thing, I just like to work with real numbers as oppose to some random variables.
Last edited on
First condition holds, 2nd condition:
T(cv) = cT(v)
cv = ca1, ca2
T(cv) = ca1 / 5, 3ca2 / 5
T(v) = a1 / 5, 3a2 / 5
cT(v) = ca1 / 5, 3ca2 / 5
Why would anyone go through all this mess if you could just work with simple 1, 2, 3 values. The result will be the same. If both conditions are met, the result is a LT, if one condition fails, the result is not a LT.
T(cv) = cT(v)
cv = ca1, ca2
T(cv) = ca1 / 5, 3ca2 / 5
T(v) = a1 / 5, 3a2 / 5
cT(v) = ca1 / 5, 3ca2 / 5
Why would anyone go through all this mess if you could just work with simple 1, 2, 3 values. The result will be the same. If both conditions are met, the result is a LT, if one condition fails, the result is not a LT.
Last edited on
It's not a matter of preference. The definition of linear transformation (https://en.wikipedia.org/wiki/Linear_map#Definition_and_first_consequences ) says that the conditions must hold for the entire domain of the function. If you're making a proof you have to do it rigorously. You can't prove that a condition holds for an entire set by proving that it holds for a single element of the set. There are many functions f such that
* f((1, 2) + (3, 4)) = f(1, 2) + f(3, 4)
* f(5 (1, 2)) = 5 f(1, 2)
that are not linear transformations.
You don't know if the result will be the same until you've proven it. That's the point of a proof.
Imagine a proof of Fermat's Last Theorem made with that kind of rigor.
2^3 + 3^3 != 5^3
6^4 + 8^4 != 100^4
6^5 + 8^5 != 100^5
Yeah, good enough. QED.
* f((1, 2) + (3, 4)) = f(1, 2) + f(3, 4)
* f(5 (1, 2)) = 5 f(1, 2)
that are not linear transformations.
| The result will be the same. |
Imagine a proof of Fermat's Last Theorem made with that kind of rigor.
2^3 + 3^3 != 5^3
6^4 + 8^4 != 100^4
6^5 + 8^5 != 100^5
Yeah, good enough. QED.
@MultiMedia, you have seen two ways to prove this. You could choose to plug in real numbers or some arbitrary variables. In my university, I used to prove these with real numbers and my professor would grant me the mark in full. Some professors may want you to prove this with arbitrary variables for the reason helios stated earlier. I think you should discuss this with your professor and see if it would be fine if you could go about it with numbers. If not, then use variables, it's the exact same pattern I showed you, just beware of the syntax.
closed account (48T7M4Gy)
| you have seen two ways to prove this |
| Why would anyone go through all this mess if you could just work with simple 1, 2, 3 values. |
Last edited on
Topic archived. No new replies allowed.