# Find out whether linearity for the functions $f$ and $g$ persists

Given $$f: \mathbb{C}^3 \rightarrow \mathbb{C}^2, \begin{bmatrix} a\\ b\\ c \end{bmatrix} \mapsto \begin{bmatrix} ia+b\\ c \end{bmatrix}, \,\,\,\,\,\,\,g: \mathbb{C}^3 \rightarrow \mathbb{C}^2, \begin{bmatrix} a\\ b\\ c \end{bmatrix} \mapsto \begin{bmatrix} ia+b\\ c+1 \end{bmatrix}$$

I like to find out if linearity exists for each function? If I understood correctly, it needs to be shown that they are homogenous and additive. So let $$\vec{v_1}=\begin{bmatrix} a_1\\ b_1\\ c_1 \end{bmatrix} \,\,\,\,, \vec{v_2}=\begin{bmatrix} a_2\\ b_2\\ c_2 \end{bmatrix} \,\,\,\, \text{ where each is from } \,\, \mathbb{C^3}$$

Because the functions need to be homogenous and additive, we need to do

$$f(\vec{v_1}+\vec{v_2}) = f(\begin{bmatrix} a_1+a_2\\ b_1+b_2\\ c_1+c_2 \end{bmatrix}) = \begin{bmatrix} i(a_1+a_2)+(b_1+b_2)\\ (c_1+c_2) \end{bmatrix}= \begin{bmatrix} (ia_1+b_1)+(ia_2+b_2)\\ (c_1+c_2) \end{bmatrix}$$

But from here I don't know how to continue and what to do ? :C

• The next step in proving homogeneity is to calculate $f(\vec{v_1})$ and $f(\vec{v_2})$, then add the results. If this is the same as $f(\vec{v_1} + \vec{v_2})$ as you have correctly calculated, then you have additivity. Jun 4 '20 at 15:34

You want to show $$f(v_1 + v_2) = f(v_1) + f(v_2)$$. If we compute the values $$f(v_1)$$ and $$f(v_2)$$ we find that \begin{align*} f(v_1) + f(v_2) &= \begin{bmatrix} ia_1 + b_1 \\ c_1 \end{bmatrix} + \begin{bmatrix} ia_2+b_2\\ c_2 \end{bmatrix} \\ &= \begin{bmatrix} (ia_1+b_1)+(ia_2+b_2)\\ c_1+c_2 \end{bmatrix} \\ &= f(v_1 + v_2). \end{align*} Thus, we see that $$f$$ is linear (if we quickly compute $$f(av) = af(v)$$). Another way to see this quickly is to note that all the components of $$f(v)$$ are linear combinations of the entries of $$v$$. This is not the case for $$g$$ due to the addition of 1 in the second component, hence $$g$$ is not linear. Alternatively, one can plug in $$0$$ to get $$g(0) = \begin{bmatrix} 0\\ 1 \end{bmatrix} \ne 0.$$
• This proves $f$ is additive. We'd need a proof of homogeneity, i.e. $f(cv) = cf(v)$ for all scalars $c$ and vectors $v$, in order to be linear. Unfortunately, proving $f(0) = 0$ doesn't cut it! Jun 4 '20 at 15:48