Linear Transformation Condition By definition, a transformation is linear if and only if
(a) $T(\vec{v}+\vec{w})=T(\vec{v})+T(\vec{w})$
(b) $T(k\vec{v}) = kT(\vec{v})$
My question is that is the second condition indispensable? For example, use (a), we have $T(\vec{x}+\vec{x})=T(\vec{x})+T(\vec{x})$, that is to say, $T(2\vec{x})= 2T(\vec{x})$. Similarly, we can induce all cases expressed by (b) from condition (a). Is condition (b) redundant? 
 A: As noted in the comments, the scalar $k$ need not be an integer, hence the two clauses of the definition are not redundant.
More generally, though, the independence of these two conditions can be more easily seen by recalling the definition of a vector space. The scalars in question can belong to any field $\Bbb F$ over which the vector space is defined. This means that you could be working with real numbers, or complex numbers, or rational functions, or matrices, or, well, anything that behaves like a field.
A: What you observe works... but only in a very limited context.
Recall that a vector space requires two pieces of information: the set of vectors with their addition, and the field of scalars and scalar multiplication (scalar times a vector).
If your field of scalars is the rational numbers (and just the rational numbers), or if it is a field of prime order $\mathbb{F}_p$, then what you are trying to do works!
(If you don’t know about fields of prime elements, then you can ignore them below)

Lemma. Let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces over $\mathbb{F}$, and let $T\colon\mathbf{V}\to\mathbf{W}$ be an additive function; that is, for all $\mathbf{x,y}\in\mathbf{V}$, $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$. Then:
  
  
*
  
*For all integers $n$ and all $\mathbf{x}\in\mathbf{V}$, $T(n\mathbf{x})=nT(\mathbf{x})$.
  
*If the integer $n$ is invertible in $\mathbb{F}$, then for all $\mathbf{x}\in\mathbf{V}$ we have $T(\frac{1}{n}\mathbf{x}) = \frac{1}{n}T(\mathbf{x})$.
  

Proof. 1. First, note that $T(\mathbf{0}) = T(\mathbf{0}+\mathbf{0}) = T(\mathbf{0})+T(\mathbf{0})$, so $T(\mathbf{0}) = \mathbf{0}$. 
Next, we first prove the result for nonnegative $n$ by induction. If $n=0$ the result follows from the above observation. If $n=1$, the result is immediate. Assume that result holds for $k$, so that $T(k\mathbf{x}) = kT(\mathbf{x})$. Then by the additivity we have:
$$T((k+1)\mathbf{x}) = T(k\mathbf{x}+\mathbf{x}) = T(k\mathbf{x})+T(\mathbf{x}) = kT(\mathbf{x})+T(\mathbf{x}) = (k+1)T(\mathbf{x}).$$
This proves the result for all nonnegative $n$. Now, if $m\lt 0$, then $n=-m\gt 0$. Then
$$\begin{align*}
\mathbf{0} &= T(\mathbf{0}) = T(0\mathbf{x})\\
&= T((n+m)\mathbf{x})\\
&= T(n\mathbf{x}+m\mathbf{x}) = T(n\mathbf{x}) + T(m\mathbf{x})\\
&= nT(\mathbf{x}) + T(m\mathbf{x}).
\end{align*}$$
Therefore, $T(m\mathbf{x}) = -nT(\mathbf{x}) = mT(\mathbf{x})$, which proves the result for $m$. Thus, this holds for all integers.


*We have
$$nT\left(\frac{1}{n}\mathbf{x}\right) = T\left(\frac{n}{n}\mathbf{x}\right) = T(\mathbf{x}),$$
and so multiplying through by $\frac{1}{n}$ we get $T(\frac{1}{n}\mathbf{x}) = \frac{1}{n}T(\mathbf{x})$, as claimed. $\Box$

Theorem. Let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces over the field $\mathbb{F}$ and let $T\colon \mathbf{V}\to\mathbf{W}$ be an additive function; that is, $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$ for all $\mathbf{x,y}\in\mathbf{V}$. 
  
  
*
  
*If $\mathbb{F}=\mathbb{Q}$, then $T$ is a linear transformation; that is, for all rational numbers $q$ and all vectors $\mathbf{x}\in\mathbf{V}$, we have $T(q\mathbf{x})=qT(\mathbf{x})$.
  
*If $\mathbb{F}=\mathbb{F}_p$ with $p$ a prime, then $T$ is a linear transformation; that is, for all $\alpha\in\mathbb{F}_p$ and all $\mathbf{x}\in\mathbf{V}$, $T(\alpha\mathbf{x})=\alpha T(\mathbf{x})$. 
  

Proof. For 1, simply write $q=\frac{r}{s}$. Then
$$T(q\mathbf{x}) = T\left(\frac{r}{s}\mathbf{x}\right) = T\left({1}{s}\left( r\mathbf{x}\right)\right) = \frac{1}{s}T(r\mathbf{x}) = \frac{r}{s}T(\mathbf{x}).$$
As for 2, it follows because $\alpha$ is just $1$ added to itself a finite number of times. $\Box$
So if your field is $\mathbb{Q}$ (or a finite field of prime order), then additivity implies homogeneity (that’s the name for the property that $T(\alpha \mathbf{x}) = \alpha T(\mathbf{x})$. 
However, if the field is not $\mathbb{Q}$ or a field of prime order, (and you assume the Axiom of Choice, or equivalently, that every vector space has a basis) then you can have a function that is additive but is not homogeneous.
“Explicitly” (modulo finding a basis), say $\mathbb{F}$ is a field that contains $\mathbb{Q}$ (that is, characteristic $0$); if this is too high falutin for you, just think $\mathbb{R}$ or $\mathbb{C}$. Let $\mathbf{V}$ be a nonzero vector space over $\mathbb{F}$. Then we can view it as a vector space over $\mathbb{Q}$. Let $\mathbf{v}\neq\mathbf{0}$ be a nonzero vector, and extend it to a basis $\mathcal{B}$ for $\mathbf{V}$ over $\mathbb{Q}$. Now define $T\colon \mathbf{V}\to\mathbf{V}$ by sending $\mathbf{v}$ to itself and every other vector in $\mathcal{B}$ to $\mathbf{0}$. This is additive (it is $\mathbb{Q}$-linear), but if $\alpha\in\mathbb{F}-\mathbb{Q}$, then $\alpha\mathbf{v}$ is not a $\mathbb{Q}$-scalar multiple of $\mathbf{v}$, and so $T(\alpha\mathbf{v})\neq \alpha\mathbf{v}$ (because $T(\mathbf{x})$ is a $\mathbb{Q}$-scalar multiple of $\mathbf{x}$ for every $\mathbf{x}$). So $T$ is additive, but cannot be homogeneous. Thus, you can find functions between $\mathbb{F}$ vector spaces that are additive but not linear.
A similar argument holds if the prime field of $\mathbb{F}$ is a finite field of prime order. 
So we have:

Theorem. Let $\mathbb{F}$ be a field. Then every additive function between $\mathbb{F}$-vector spaces if $\mathbb{F}$-linear if and only if $\mathbb{F}$ is $\mathbb{Q}$ or a field of prime order (that is, if and only if $\mathbb{F}$ is a prime field). 

On the other hand, there is no field for which homogeneity will always imply additivity.
Theorem. Let $\mathbf{V}$ be a vector space. If $\mathbf{V}$ has dimension at least $2$, then there is a function $\mathbf{V}\to\mathbf{V}$ that is homogeneous but not additive.
Proof. Let $\mathbf{v}$ and $\mathbf{w}$ be linearly independent, and complete this to a basis $\mathcal{B}$. Define $T\colon\mathbf{V}\to\mathbf{V}$ as follows: if $\mathbf{x}$ is a scalar multiple of $\mathbf{v}$ or is a scalar multiple of $\mathbf{w}$, map it to itself. Otherwise, map $\mathbf{x}$ to $\mathbf{0}$.
This is clearly not additive: for $T(\mathbf{v}) = \mathbf{v}$, and $T(\mathbf{w}) = \mathbf{w}$, but $T(\mathbf{v}+\mathbf{w}) = \mathbf{0}\neq \mathbf{v}+\mathbf{w}$. 
However, $T$ is homogeneous: if $\alpha$ is the zero scalar, then we certainly have $T(\alpha \mathbf{x}) = \alpha T(\mathbf{x})$; if $\alpha\neq 0$, then consider the three cases: if $\mathbf{x}$ is a scalar multiple of $\mathbf{v}$, then so is $\alpha\mathbf{x}$, so $T(\alpha\mathbf{x}) = \alpha\mathbf{x}=\alpha T(\mathbf{x})$. Similarly if $\mathbf{x}$ is a multiple of $\mathbf{w}$. But if $\mathbf{x}$ is not a multiple of $\mathbf{v}$ nor of $\mathbf{w}$, then neither is $\alpha\mathbf{x}$ (since $\alpha\neq 0$), to $T(\alpha\mathbf{x}) = \mathbf{0} = \alpha\mathbf{0} = \alpha T(\mathbf{x})$. So $T$ is homogeneous, but not additive. $\Box$
I leave it to you to show that if $\dim(\mathbf{V})=1$, then any homogeneous function is necessarily additive and hence linear. 
