0
$\begingroup$

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

$\endgroup$
1
  • 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
1
$\begingroup$

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 $$g(0) = \begin{bmatrix} 0\\ 1 \end{bmatrix} \ne 0.$$

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.