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I am requested to determine wether these functions are linear or not; to do that, I've to verify both the necessary conditions that are:

$f(x+y) = f(x) + f(y)$

$f(\alpha x) = \alpha f(x)$

Now, my functions are:

$f(x_1,x_2,x_3) = (x_1+x_2,x_3)$ f:R3→R2

$f(x_1,x_2) = (x_2,-x_1,x_1+3x_2)$ f:R2→R3

How can I proceed??

Thank you, in advance.

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    $\begingroup$ Hint: consider $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ in the first case, and $x=(x_1,x_2)$ and $y=(y_1,y_2)$. It helps seeing the situation clearly. $\endgroup$
    – Martigan
    Commented Nov 21, 2014 at 14:38
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    $\begingroup$ Say what are the domains of these functions? I guess the first one is $f : \mathbb{R}^3 \rightarrow \mathbb{R}^2$, but make it clear! $\endgroup$
    – brick
    Commented Nov 21, 2014 at 14:38
  • $\begingroup$ Yes, sorry.. the domain is what you said in the first one.. $\endgroup$
    – RedViper16
    Commented Nov 21, 2014 at 14:42
  • $\begingroup$ Also, for the second one: $f : \mathbb{R}^2 \rightarrow \mathbb{R}^3$. Make it clear in the question. $\endgroup$ Commented Nov 21, 2014 at 14:49
  • $\begingroup$ Sorry...I've corrected now.. but someone can help me solve it? $\endgroup$
    – RedViper16
    Commented Nov 21, 2014 at 14:52

4 Answers 4

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For the first one, remembering that by definition for $a = (a_1, a_2, \dots, a_n)$ and $b = (b_1, b_2, \dots, b_n)$ addition is defined by $a+b = (a_1, a_2, \dots, a_n) + (b_1, b_2, \dots, b_n) = (a_1 + b_1, a_2 + b_2, \dots, a_n + b_n)$ and that multiplication by a scalar is defined as $\alpha a = (\alpha a_1, \alpha a_2, \dots, \alpha a_n)$ for any scalar $\alpha$.

We may shorten any proof attempt by proving the statement: for all $x, y \in \mathscr{D}(f)$ and all $\alpha , \beta \in \mathbb{C}$ we have $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$ (since setting alpha and beta equal to 1 gives the first condition and setting beta equal to zero gives the second condition).

Consider for the first one, which was defined as $f( (x_1, x_2, x_3) ) = (x_1 + x_2, x_3)$ for the points $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$ and arbitrary scalars $\alpha, \beta$.

Then $f(\alpha x + \beta y) = f( (\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \alpha x_3 + \beta y_3)) = (\alpha x_1 + \beta y_1 + \alpha x_2 + \beta y_2, \alpha x_3 + \beta y_3)~~~~$

here i just used the definition of addition of vectors and scalar multiplication and the definition of the function itself

$ = (\alpha( x_1 + x_2) + \beta (y_1 + y_2), \alpha x_3 + \beta y_3) ~~~$*grouped x's and y's together and factored out scalar constants*

$=\alpha (x_1 + x_2, x_3) + \beta (y_1 + y_2, y_3) = \alpha f((x_1, x_2, x_3)) + \beta f((y_1, y_2, y_3)) = \alpha f(x) + \beta f(y)~~~~$ seperated vector into addition of two vectors and factored out scalar constant, then used definition of our function to complete

Therefore, the first function is indeed linear. Try it now with the second function yourself and see where you can get.

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Let $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3)$. Then $x+y=(x_1+y_1,x_2+y_2,x_3+y_3)$ and

\begin{align} f(x+y) &=f(x_1+y_1,x_2+y_2,x_3+y_3)\\ &=(x_1+y_1+x_2+y_2,x_3+y_3)\\ &=(x_1+x_2,x_3)+(y_1+y_2,y_3)\\ &=f(x)+f(y). \end{align} So $f$ is linear. Others can be shown similarly.

EDIT $f$ is additive.

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  • $\begingroup$ That shows the first property of linearity only. Left out scalar multiplication which is also important. $\endgroup$
    – JMoravitz
    Commented Nov 21, 2014 at 14:56
  • $\begingroup$ \begin{align}f(\alpha x) &=f(\alpha x_1,\alpha x_2,\alpha x_3)\\ &=(\alpha x_1+\alpha x_2,\alpha x_3)\\ &=\alpha (x_1+x_2,x_3)\\ &=\alpha f(x)\end{align} $\endgroup$
    – MBYagbasan
    Commented Nov 21, 2014 at 15:02
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First one : $f((x_1, x_2, x_3) + (y_1, y_2, y_3)) = f((x_1 + y_1, x_2 + y_2, x_3 + y_3) = (x_1 + y_1 + x_2 + y_2, x_3 + y_3) = (x_1 + x_2, x_3) + (y_1 + y_2, y_3) = f((x_1, x_2, x_3)) + f((y_1, y_2, y_3))$

$f(a(x_1, x_2, x_3)) = f((ax_1, ax_2, ax_3)) = (ax_1 + ax_2, ax_3) = (a(x_1 + x_2), ax_3) = a(x_1 + x_2, x_3) = af((x_1, x_2, x_3))$

Second one : $f((x_1, x_2) + (y_1, y_2)) = f((x_1 + y_1, x_2 + y_2)) = (x_2 + y_2 - x_1 - y_1, x_1 + y_1 + 3x_2 + 3y_2) = (x_2 - x_1, x_1 + 3x_2) + (y_2 - y_1, y_1 + 3y_2) = f((x_1, x_2)) + f((y_1, y_2))$

$f(a(x_1, x_2)) = f((ax_1, ax_2)) = (ax_2 - ax_1, ax_1 + 3ax_2) = a(x_2 - x_1, x_1 + 3x_2) = af((x_1, x_2))$

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Here's a trick; if your functions are linear, then they are obtained by $$ f(x)=Ax $$ for some matrix $A$, where vectors are written as columns.

The matrix can also be written explicitly: its columns are exactly $f(e_1)$, $f(e_2)$ and $f(e_3)$ (in the first case where the domain is $\mathbb{R}^3$) where $e_1$, $e_2$ and $e_3$ are the vectors in the standard (or canonical) basis.

Since, by definition, $$ f(e_1)=f\left(\begin{bmatrix}1\\0\\0\end{bmatrix}\right)= \begin{bmatrix}1\\0\end{bmatrix}, \qquad f(e_2)=\begin{bmatrix}1\\0\end{bmatrix}, \qquad f(e_3)=\begin{bmatrix}0\\1\end{bmatrix}, $$ the matrix is $$ A=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Now, for a generic vector, $$ Ax=A\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}= \begin{bmatrix} x_1+x_2\\ x_3 \end{bmatrix}=f(x) $$ and this ends the verification.

Try your hand with the second map.

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