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

  • 1
    $\begingroup$ 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. $\endgroup$
    – user793679
    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 $$g(0) = \begin{bmatrix} 0\\ 1 \end{bmatrix} \ne 0.$$

  • $\begingroup$ 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! $\endgroup$
    – user793679
    Jun 4 '20 at 15:48
  • $\begingroup$ Sorry, I mixed up what I wanted to write. Should be fixed now; I think OP can do the computation on their own now that they have an example of how it works :) $\endgroup$ Jun 4 '20 at 15:55
  • $\begingroup$ Many thanks and also for the comment this is very helpful already! x) $\endgroup$
    – kathelk
    Jun 4 '20 at 16:03

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