# Show that $f$ is a linear transformation

Let $f: \mathbb{R}^m \to \mathbb{R}^n$, where $f(tx)=tf(x)$, for all $x \in \mathbb{R}^m$ and $t \in \mathbb{R}$. Show that $f$ is a linear transformation.

-
What have you tried? Do you know the definition of a linear transformation? Also, are you sure that this is the full question? –  KReiser Nov 5 '12 at 15:58
Please do not post unnecessary comments! thank you –  Jarbas Dantas Silva Nov 5 '12 at 16:03
I think you will find that KReiser is actually trying to be helpful, here. If we know what you have tried we have a better chance of being able to help. It is also important that we know what definition of linear you are using, as there is a chance that the result you are trying to prove is not actually true! –  Old John Nov 5 '12 at 16:06
I think that the result it's not true , unless $m=1$. I think that $f:\Bbb R^2 \to \Bbb R$ that $f(te_1)=t$ for $t\in \Bbb R$ and $f(x)=0$ otherwise, it's well defined. And it's a counterexample –  Daniel Nov 5 '12 at 16:12
Is there a (homework or whatever) exercise you want to solve? In that case you should put the whole exercise here. –  Berci Nov 5 '12 at 17:23

What you have provided is not enough to show that $f$ is a linear transformation. It shows that $f$ preserves scalar multiplication but does not show that $f$ preserves addition. These two properties are independent of each other.

A non-trivial example is as follows. For each ray $\ell$ through the origin in $\mathbb{R}^{n}$, choose a scalar $\alpha_{\ell} \in \mathbb{R}$. Then define $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ as follows: $$\forall \mathbf{x} \in \mathbb{R}^{n}: \quad \mathbf{x} \in \ell \quad \Longrightarrow \quad f(\mathbf{x}) \stackrel{\text{def}}{=} \alpha_{\ell} \cdot \mathbf{x}.$$ It is seen that $f$ scales each ray passing through the origin in a manner that depends only on the ray. Hence, by choosing the $\alpha_{\ell}$'s carefully, one can make $f$ non-linear.

Here is how we can do it. Pick $\mathbf{x},\mathbf{y} \in \mathbb{R}_{n} \setminus \{ \mathbf{0} \}$ such that $\mathbf{x} \notin \text{Span}(\{ \mathbf{y} \})$. Let $\ell(\mathbf{x})$ and $\ell(\mathbf{y})$ be the unique rays passing through $\mathbf{x}$ and $\mathbf{y}$ respectively and also the origin. Clearly, $\ell(\mathbf{x}) \neq \ell(\mathbf{y})$. Set $\alpha_{\ell(\mathbf{x})} = \alpha_{\ell(\mathbf{y})} = 1$. Observe that $\mathbf{x} + \mathbf{y} \neq \mathbf{0}$ and that $\mathbf{x} + \mathbf{y}$ belongs to a unique ray $\ell(\mathbf{x} + \mathbf{y})$ passing through the origin that is identical to neither $\ell(\mathbf{x})$ nor $\ell(\mathbf{y})$. Now, if $f$ were linear, we would expect to have \begin{align} f(\mathbf{x} + \mathbf{y}) &= f(\mathbf{x}) + f(\mathbf{y}) \\ &= \alpha_{\ell(\mathbf{x})} \cdot \mathbf{x} + \alpha_{\ell(\mathbf{y})} \cdot \mathbf{y} \\ &= 1 \cdot \mathbf{x} + 1 \cdot \mathbf{y} \\ &= \mathbf{x} + \mathbf{y} \\ &\neq \mathbf{0}. \end{align} However, as $\ell(\mathbf{x} + \mathbf{y}) \neq \ell(\mathbf{x}),\ell(\mathbf{y})$, we are free to choose $\alpha_{\ell(\mathbf{x} + \mathbf{y})} = 0$, which yields $f(\mathbf{x} + \mathbf{y}) = \mathbf{0}$ instead. Therefore, $f$ cannot be linear.

-

The correct definition is as follows:

Def: $f$ is linear if, as you write, $f(tx)=tf(x)$ for all vector $x$ and scalar $t$,

and $f(x+y)=f(x)+f(y)$ for all vectors $x,y$.

Your question as you stated is not true for $m>1$ (basically because it misses a part of the definition), and a counterexample is given in the comments by Daniel. Please show concrete example in $\Bbb R^2$ for $x,y$ such that $f(x+y)\ne f(x)+f(y)$ for that particular $f$.

-