# Showing a transformation is not linear

1. Jan 28, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

http://img526.imageshack.us/img526/4926/screenshot20120128at941.png [Broken]

3. 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?

Last edited by a moderator: May 5, 2017
2. Jan 28, 2012

### micromass

This is perfect!!

3. Jan 28, 2012

### 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 [Broken]

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.

Last edited by a moderator: May 5, 2017
4. Jan 28, 2012

### micromass

Also good!!

5. Jan 28, 2012

### TranscendArcu

http://img192.imageshack.us/img192/7066/screenshot20120128at100.png [Broken]

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.

Last edited by a moderator: May 5, 2017
6. Jan 28, 2012

### micromass

Fine!!

7. Jan 28, 2012

### TranscendArcu

http://img85.imageshack.us/img85/6540/screenshot20120128at101.png [Broken]
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.

Last edited by a moderator: May 5, 2017
8. Jan 28, 2012

### micromass

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

9. Jan 28, 2012

### 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 [Broken]

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.

Last edited by a moderator: May 5, 2017
10. Jan 28, 2012

### 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??

11. Jan 28, 2012

### TranscendArcu

:,(
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.

12. Jan 28, 2012

### micromass

Tht's good!!

13. Jan 29, 2012

### 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 [Broken]

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.

Last edited by a moderator: May 5, 2017
14. Jan 29, 2012

Seems ok!!