Showing a transformation is not linear

[TranscendArcu]

Homework Statement

http://img526.imageshack.us/img526/4926/screenshot20120128at941.png

The Attempt at a Solution

A linear transformation must satisfy the property: T(a \vec{X} ) = a T(\vec{X} ) \forall X \in V, a \in R. However, it is not in general true that (a^2 x^2,a^2 y^2 = a(x^2,y^2). In particular, we can see this fails for a = 3, x= 4,y=5. Indeed,

(3^2 4^2,3^2 5^2) ≠ 3(4^2,5^2)
(9 \ast 16,9 \ast 25) ≠ 3(16,25)
(144,225) ≠ (48,75)

Thus, this is not a linear transformation.

Am I doing this right?

 

[micromass]

This is perfect!

 

[TranscendArcu]

Good to hear I did the first one right. Thanks! There are four of these problems in total and I want to make sure I know how to do them correctly.

http://img267.imageshack.us/img267/844/screenshot20120128at956.png

A linear transformation must satisfy the property that T(\vec{A} + \vec{B}) = T(\vec{A}) + T(\vec{B}). Let two vectors in R^3 be denoted by (x_1,y_1,z_1),(x_2,y_2,z_2). Thus,

T((x_1,y_1,z_1) + (x_2,y_2,z_2)) = T(x_1 +x_2,y_1 + y_2, z_1 + z_2)
T(x_1 +x_2,y_1 + y_2, z_1 + z_2) = (x_1 +x_2+y_1 + y_2+z_1 + z_2,1)

But, T(x_1,y_1,z_1) + T(x_2,y_2,z_2) = (x_1+y_1+z_1,1) + (x_2+y_2+z_2,1)
(x_1+y_1+z_1,1) + (x_2+y_2+z_2,1) = (x_1 +x_2+y_1 + y_2+z_1 + z_2,2)

However, (x_1 +x_2+y_1 + y_2+z_1 + z_2,1) ≠ (x_1 +x_2+y_1 + y_2+z_1 + z_2,2). Thus, this is not a linear transformation.

 

[micromass]

Also good!

 

[TranscendArcu]

http://img192.imageshack.us/img192/7066/screenshot20120128at100.png

We said earlier that a linear transformation has the property: T(a \vec{X} ) = a T(\vec{X}). In this problem,

T(a x) = (1,-1). However, a T(x) = a (1,-1) = (a,-a).

The equation (1,-1) = (a,-a) holds only in the case a =1. Because the equation does not hold for any other number, this is not a linear transformation.

 

[micromass]

Fine!

 

[TranscendArcu]

http://img85.imageshack.us/img85/6540/screenshot20120128at101.png
We'll use the same property as in the first and third problems to show this is not a linear transformation.

T[a(x,y)] = T(ax,ay) = (a^2 x y,ay,ax). However, aT(x,y) = a(xy,y,x) = (axy,ay,ax). Again, we observe that (a^2 x y,ay,ax) = (axy,ay,ax) will be true only in two particular cases. These are a = 1,a= 0. Because the equation is not true in general, we conclude this is not a linear transformation.

 

[micromass]

Good! (maybe it's also good to find explicit values of a,x and y such that the equation doesn't hold)

 

[TranscendArcu]

Okay. I'll add a specific case to show that the equation does not hold. Thanks! My other homework problems are a little bit different, but I'd like to talk about this one in particular:

http://img220.imageshack.us/img220/7180/screenshot20120128at104.png

First, I considered the idea of multiplication as a binary operator. That is, for well-chosen a_1,a_2,a_3 \in R it is conceivable that a_1 * a_2 = a_3. That is, I am able to input two numbers and get one number as a result. With regard to this question, I said let T(x) = y where y \in R. Thus, using the idea of multiplication as a binary operator, I would have x * t = y. In the cases that x ≠ 0 I can solve to find t = \frac{y}{x} \in R. In the case x=0, we know that any linear transformation of the zero vector of the domain must map to the zero vector of the codomain. In this case, t can be any such number because t * \vec{0} = \vec {0}.

First of all, is this a proof? It seems conceivable to me that this could work but I'm not entirely sure.

 

[micromass]

Hmmm, I don't quite understand the proof :( So I guess it's not correct.

Here's a way to prove it: let T(1)=t. Can you now use linearity to show that T(x)=tx??

 

[TranscendArcu]

Hmmm, I don't quite understand the proof :( So I guess it's not correct.

:,(

Here's a way to prove it: let T(1)=t. Can you now use linearity to show that T(x)=tx??

Alright, letting T(1) = t. We know that x \in R. Thus, we may treat x as a kind of scalar. We can write,

T(x) = T(1*x) = xT(1) = xt. Thus, the existence of such a t is shown.

 

[micromass]

:,(
Alright, letting T(1) = t. We know that x \in R. Thus, we may treat x as a kind of scalar. We can write,

T(x) = T(1*x) = xT(1) = xt. Thus, the existence of such a t is shown.

Tht's good!

 

[TranscendArcu]

Okay. That's much shorter than what I had previously, so that's an improvement (and it's right, too!). Below is another problem that I'd like my work checked on.

http://img337.imageshack.us/img337/1043/screenshot20120129at101.png

Let \left\{ A_1,...,A_n \right\} be a basis for V, and \left\{ B_1,...,B_n \right\} be a basis for W. If T is injective, then T(A_i) ≠ T(A_k) where 1≤i,k≤n. If T is injective, then ker(T) = \left\{ 0 \right\}. We know, dim(V) = dim(ker(T)) + dim(Im(T)). We write n = 0 + dim(Im(T)). This implies dim(Im(T)) = n, and thus, Im(T) = V, which means T is surjective.

Suppose T is surjective. Then Im(T) = V. Thus, dim(Im(T)) = dim(V). We write, dim(Im(T)) = n. We know, dim(V) = dim(ker(T)) + dim(Im(T)). We write, n = dim(ker(T)) + n. This implies dim(ker(T)) = 0, and thus ker(T) = \left\{ 0 \right\}. This means that T is injective.

 

[micromass]

Seems ok!