Wikipedia states:

Let V and W be vector spaces over the same field K. A function f : V → W is said to be a linear map if for any two vectors x and y in V and any scalar α in K, the following two conditions are satisfied:

(L1) $f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!$ additivity

(L2) $f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \! $ homogeneity of degree 1

This is equivalent to requiring the same for any linear combination of vectors, i.e. that for any vectors x1, ..., xm ∈ V and scalars a1, ..., am ∈ K, the following equality holds:

(L3) $f(a_1 \mathbf{x}_1+\cdots+a_m \mathbf{x}_m) = a_1 f(\mathbf{x}_1)+\cdots+a_m f(\mathbf{x}_m). \!$

But in a lot of sources it is stated that if we can show

(L4) $f(b \mathbf{x}_1+\mathbf{x}_2) = b \cdot f(\mathbf{x}_1)+f(\mathbf{x}_2). \!$ for all x1, x2 ∈ V and for all b ∈ V.

this will imply (L1) and (L2). Is this correct, or do we have to show: (which I think is equivalent to (L3))

(L5) $f(b_1 \mathbf{x}_1+ b_2 \mathbf{x}_2) = b_1 \cdot f(\mathbf{x}_1)+b_2 \cdot f(\mathbf{x}_2). \!$ for all x1, x2 ∈ V and for all b ∈ K.

Which one will imply (L1) and (L2)? (L3),(L4),(L5)?

  • 1
    $\begingroup$ In (L4), set $b=1$ to get (L1), and $x_2=0$ to get (L2). Similar methods work for (L3) and (L5). $\endgroup$
    – roman
    Mar 11, 2015 at 8:49
  • 1
    $\begingroup$ In my opinion, characterisation (L4), albeit correct from a logical point of view, is quite unnatural: why should variable $\mathbf x_1$ play a special role w.r.t. $\mathbf x_2$? $\endgroup$
    – Bernard
    Mar 11, 2015 at 9:13
  • $\begingroup$ I think it does not, since it holds for all $x_1$ and $x_2$. But the point is, that (L4) shortens the proof over (L5) that $f$ is a linear map. $\endgroup$
    – NoMorePen
    Mar 11, 2015 at 9:18

1 Answer 1


L5 is a consquence of L4. To prove this statement first prove that from L4 whe have: $$ f(0)=f(-1 \cdot x +x)=-f(x)+f(x)=0 \qquad \forall x $$ then prove L5: $$ f(b_1x_1+b_2x_2)=b_1f(x_1)+f(b_2x_2)=b_1f(x_1)+f(b_2x_2+0)= $$ $$ =b_1f(x_1)+b_2f(x_2)+f(0)=b_1f(x_1)+b_2f(x_2) $$ and obviously L5$\Rightarrow$ L4.


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