Linear Transformations

**vecoma** · March 30th 2011, 10:12 PM

I'm a bit confused on how to prove if a transformation is linear or not.

I know that there are 2 rules:
T(u+v) = T(u)+T(v)

and

T(ku)= kT(u)

However, I'm still not sure though how to write out the problems out in their form...Can someone explain to me step by step or give me an example?

I'm trying to learn so i can figure out the following problem. Also i tried attempting these questions and i ruled out that A is not linear. In addition, i'm not sure.. but i think C,D,E are linear. Can someone check my answers?

http://img8.imageshack.us/img8/2470/mathprobl.jpg

Thanks Alot!

**Chris L T521** · March 30th 2011, 10:34 PM

Originally Posted by vecoma

I'm a bit confused on how to prove if a transformation is linear or not.

I know that there are 2 rules:
T(u+v) = T(u)+T(v)

and

T(ku)= kT(u)

However, I'm still not sure though how to write out the problems out in their form...Can someone explain to me step by step or give me an example?

I'm trying to learn so i can figure out the following problem. Also i tried attempting these questions and i ruled out that A is not linear. In addition, i'm not sure.. but i think C,D,E are linear. Can someone check my answers?

http://img8.imageshack.us/img8/2470/mathprobl.jpg

Thanks Alot!

Well you're correct so far: A is not a linear transformation; C, D, and E are linear transformations. (What about B and F?)

To see why A isn't a linear transformation, suppose we have two vectors: $\begin{bmatrix}x_1\\x_2\end{bmatrix}$ , $\begin{bmatrix}x_1^{\prime}\\x_2^{\prime}\end{bmat rix}$

We see that

$\begin{aligned}T\left(\begin{bmatrix}x_1\\x_2\end{ bmatrix}+\begin{bmatrix}x_1^{\prime}\\x_2^{\prime} \end{bmatrix}\right) & = T\left(\begin{bmatrix}x_1+x_1^{\prime}\\x_2+x_2^{\ prime}\end{bmatrix}\right)\\ &= \begin{bmatrix}0\\(x_1+x_1^{\prime})(x_2+x_2^{\pri me})\end{bmatrix}\end{aligned}$

But $\begin{bmatrix}0\\(x_1+x_1^{\prime})(x_2+x_2^{\pri me})\end{bmatrix}\neq \begin{bmatrix}0\\x_1+x_1^{\prime}\end{bmatrix}+\b egin{bmatrix}0\\ x_2+x_2^{\prime}\end{bmatrix}= T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) +T\left(\begin{bmatrix}x_1^{\prime}\\x_2^{\prime}\ end{bmatrix}\right)$

So $T$ is not a linear transformation.

You could have also seen this with the scalar multiplication:

$\begin{aligned}T\left(k\begin{bmatrix}x_1\\x_2\end {bmatrix}\right) & = T\left(\begin{bmatrix}kx_1\\kx_2\end{bmatrix}\righ t)\\ &= \begin{bmatrix}0\\ (kx_1)(kx_2)\end{bmatrix}\\ &= \begin{bmatrix}0\\k^2x_1x_2\end{bmatrix}\\ &= k^2\begin{bmatrix}0\\x_1x_2\end{bmatrix}\\&=k^2T\l eft(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) \neq k T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) \end{aligned}$

So we see this way as well that T is not a linear transformation.

-------

To see why D is a linear transformation, suppose we have two vectors: $\begin{bmatrix}x_1\\x_2\end{bmatrix}$ , $\begin{bmatrix}x_1^{\prime}\\x_2^{\prime}\end{bmat rix}$

We see that

$\begin{aligned}T\left(\begin{bmatrix}x_1\\x_2\end{ bmatrix}+\begin{bmatrix}x_1^{\prime}\\x_2^{\prime} \end{bmatrix}\right) & = T\left(\begin{bmatrix}x_1+x_1^{\prime}\\x_2+x_2^{\ prime}\end{bmatrix}\right)\\ &= \begin{bmatrix}0\\2(x_2+x_2^{\prime})\end{bmatrix} \\ &= \begin{bmatrix}0\\2x_2+2x_2^{\prime}\end{bmatrix}\ \ &= \begin{bmatrix}0\\2x_2\end{bmatrix}+\begin{bmatrix }0\\2x_2^{\prime}\end{bmatrix}\\ &= T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) +T\left(\begin{bmatrix}x_1^{\prime}\\x_2^{\prime}\ end{bmatrix}\right)\end{aligned}$

and we also see that

$\begin{aligned}T\left(k\begin{bmatrix}x_1\\x_2\end {bmatrix}\right) & = T\left(\begin{bmatrix}kx_1\\kx_2\end{bmatrix}\righ t)\\ &= \begin{bmatrix}0\\ 2(kx_2)\end{bmatrix}\\ &= \begin{bmatrix}0\\k(2x_2)\end{bmatrix}\\ &= k\begin{bmatrix}0\\2x_2\end{bmatrix}\\ &= k T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) \end{aligned}$

Therefore, $T$ is a linear transformation.

-------

I hope this makes more sense now. Hopefully you'll be able to check the others on your own now.

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