ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: /0 on February 17, 2010, 01:52:09 am

Title: Linear transformation
Post by: /0 on February 17, 2010, 01:52:09 am: Just a quickie...

Show that the mapping T defined by $T(x_1,x_2) = (2x_1-3x_2,x_1+4,5x_2)$ is not linear.

Does the notation $T(x_1,x_2)$ mean the same thing as $T\left(\left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right]\right)$ ?

And so would this be the way to go about it?

$T\left(\left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right]\right)+T\left(\left[\begin{matrix} y_1 \\ y_2 \end{matrix}\right]\right)=\left[\begin{matrix} 2x_1-3x_2 \\ x_1+4 \\ 5x_2\end{matrix}\right]+\left[\begin{matrix} 2y_1-3y_2 \\ y_1+4 \\ 5y_2 \end{matrix}\right] \neq \left[\begin{matrix} 2(x_1+y_1)-3(x_2+y_2) \\ (x_1+y_1)+4 \\ 5(x_2+y_2) \end{matrix}\right] =T\left(\left[\begin{matrix}x_1 \\ x_2 \end{matrix}\right]+\left[\begin{matrix} y_1 \\ y_2\end{matrix}\right]\right)$

Thanks!
Title: Re: Linear transformation
Post by: Mao on February 17, 2010, 02:02:13 am: Yes. The notation is essentially the same, just different ways of representing 'vectors' (for lack of a better expression).
Title: Re: Linear transformation
Post by: /0 on February 17, 2010, 02:05:46 am: thx mao!
also the column vectors friggin pwn the ( ) vectors
Title: Re: Linear transformation
Post by: mark_alec on February 17, 2010, 02:07:41 am: Easier way: show that (0,0) does not map to (0,0,0).
Title: Re: Linear transformation
Post by: /0 on February 17, 2010, 02:10:12 am: Quote from: mark_alec on February 17, 2010, 02:07:41 am
Easier way: show that (0,0) does not map to (0,0,0).

Does that always prove it is a linear transformation? How come?
Title: Re: Linear transformation
Post by: Mao on February 17, 2010, 02:19:36 am: Quote from: /0 on February 17, 2010, 02:10:12 am
Quote from: mark_alec on February 17, 2010, 02:07:41 am
Easier way: show that (0,0) does not map to (0,0,0).

Does that always prove it is a linear transformation? How come?

If $T(\textbf{0}) \neq \textbf{0}$

Then $T(\textbf{0}) + T(\textbf{0}) \neq T(\textbf{0} + \textbf{0})$

Cbf trying to see if this always proves if something is a linear transformation, but it definitely disproves it.

EDIT: it's not 'iff', T can be a non-linear transformation and still satisfies $T(\textbf{0}) = \textbf{0}$ , such as $T(x,y) = (x^2,y^2)$
Title: Re: Linear transformation
Post by: /0 on February 17, 2010, 02:24:18 am: Ah ok, thanks
Title: Re: Linear transformation
Post by: mark_alec on February 17, 2010, 02:37:42 am: Quote from: /0 on February 17, 2010, 02:10:12 am
Quote from: mark_alec on February 17, 2010, 02:07:41 am
Easier way: show that (0,0) does not map to (0,0,0).
Does that always prove it is a linear transformation? How come?
One of the properties of a linear transformation that must be satisfied is that the 0 maps onto 0, so it is often a trivial way of showing that something is *not* a linear transformation.
Title: Re: Linear transformation
Post by: QuantumJG on February 17, 2010, 12:52:15 pm: Quote from: /0 on February 17, 2010, 02:10:12 am
Quote from: mark_alec on February 17, 2010, 02:07:41 am
Easier way: show that (0,0) does not map to (0,0,0).

Does that always prove it is a linear transformation? How come?

I don't exactly know why it is so, but I know I'm REALLY rusty with linear algebra.

Quote from: Mao on February 17, 2010, 02:19:36 am
Quote from: /0 on February 17, 2010, 02:10:12 am
Quote from: mark_alec on February 17, 2010, 02:07:41 am
Easier way: show that (0,0) does not map to (0,0,0).

Does that always prove it is a linear transformation? How come?

If $T(\textbf{0}) \neq \textbf{0}$

Then $T(\textbf{0}) + T(\textbf{0}) \neq T(\textbf{0} + \textbf{0})$

Cbf trying to see if this always proves if something is a linear transformation, but it definitely disproves it.

EDIT: it's not 'iff', T can be a non-linear transformation and still satisfies $T(\textbf{0}) = \textbf{0}$ , such as $T(x,y) = (x^2,y^2)$

As what mao said, it doesn't always tell us the transformation is linear, but if this condition is not satisfied then no matter what the transformation is not linear.
Title: Re: Linear transformation
Post by: mark_alec on February 17, 2010, 03:10:01 pm: If $T$ is a linear transformation then the following are true:
1. $T(x+y) = T(x) + T(y)$
2. $T(ax) = aT(x)$

The zero vector mapping to the zero vector is a consequence of this. If you can find a counter-example to any of these properties then $T$ is not a linear transformation.