I think that it can be A and E. How can I recognize this? If someone might show me one I can figure the end the rest.

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YushaYusha
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To test whether T is a straight transformation, you require to check that for some vectors $a$ and $b$ and also some continuous $c$$T(a + b) = T(a) +T(b)$$ $$T(ca) = cT(a)$$ $$T(0) = 0$$So for example,A.$T(x_1,x_2,x_3)=(x_1,0,x_3)$$T(x_1+y_1,x_2+y_2,x_3+y_3)=(x_1+y_1,0(x_2+y_2),x_3+y_3)=T(x_1,0,x_3)+T(y_1,0,y_2)$$$$T(cx_1,cx_2,cx_3)=T(cx_1,(c)0,cx_3)=cT(x_1,0,x_3)$$$$T(0,0,0)=0$$B. $T(x_1,x_2)=(2x_1−3x_2,x_1+4,5x_2)$$T(x_1+y_1,x_2+y_2)=(2(x_1+y_1)−3(x_2+y_2),(x_1+y_1)+4,5(x_2+y_2))=(2x_1+2y_1−3x_2-3y_2,x_1+y_1+4,5x_2+5y_2)$$$$T(x_1,x_2)+T(y_1,y_2)=(2x_1−3x_2,x_1+4,5x_2)+(2y_1−3y_2,y_1+4,5y_2)=(2x_1−3x_2+2y_1-3y_2,x_1+y_1+8,5x_2+5y_2)\not=T(x_1+y_1,x_2+y_2)$$So B is not a linear transformation. re-superstructure point out follow edited Jun 20 "16 in ~ 21:02 answered Jun 20 "16 at 20:26 BridgetBridget 20611 silver- badge55 bronze badges$\endgroup$6 | show 1 much more comment 5$\begingroup$For$T$to it is in a linear transformation, 2 criteria must be satisfied:$T(\juniorg8.combfx+\juniorg8.combfy) = T(\juniorg8.combfx)+T(\juniorg8.combfy)$$T(a\juniorg8.combfx) = aT(\juniorg8.combfx) for a a scalar/constant. As an example, suppose \juniorg8.combfx = (x_1, x_2, x_3) and \juniorg8.combfy = (y_1, y_2, y_3). Let"s start with A. Then, T(\juniorg8.combfx) = (x_1, 0, x_3) and T(\juniorg8.combfy) = (y_1, 0, y_3). Currently$$\juniorg8.combfx+\juniorg8.combfy = (x_1 + y_1, x_2 + y_2, x_3 + y_3)\text.$$It complies with that T(\juniorg8.combfx+\juniorg8.combfy) = (x_1 + y_1, 0, x_3+y_3). Is T(\juniorg8.combfx+\juniorg8.combfy) = T(\juniorg8.combfx)+T(\juniorg8.combfy)? Now, because that a continuous a, a\juniorg8.combfx = (ax_1, ax_2, ax_3). We have actually T(a\juniorg8.combfx) = (ax_1, 0, ax_3). Furthermore, we know T(\juniorg8.combfx) = (x_1, 0, x_3), therefore aT(\juniorg8.combfx) = (ax_1, 0, ax_3). Is T(a\juniorg8.combfx) = aT(\juniorg8.combfx)? Repeat this for every one of the various other T. re-superstructure mention follow reply Jun 20 "16 in ~ 20:31 ClarinetistClarinetist 17.4k66 yellow badges5050 silver badges107107 bronze badges \endgroup add a comment | 1 \begingroup A.$$T(a\vecx)=T(ax_1, ax_2, ax_3) = (ax_1, 0, ax_3) = a(x_1, 0, x_3) = aT(\vecx)T(\vecx +\vecy)=T(x_1+y_1, x_2+y_2, x_3+y_3)\\=(x_1+y_1,0,x_3+y_3)=(x_1,0,x_3)+(y_1,0,y_3)=T(\vecx)+T(\vecy)$$Therefore the an initial transformation is direct as you effectively guessed. Just repeat the exact same procedure for B-E and also see if it works or not. re-superstructure mention monitor answered Jun 20 "16 in ~ 20:29 Chill2MachtChill2Macht 18.6k99 yellow badges3636 silver badges110110 bronze title \endgroup include a comment | 1 \begingroup A linear transformation has this defintion. T(\juniorg8.combf x+\juniorg8.combf y) = T(\juniorg8.combf x) + T(\juniorg8.combf y)\\T(c\juniorg8.combf x) = cT(\juniorg8.combf x) Show the this is (or isn"t) true because that each of the above. re-publishing cite monitor edited Jun 20 "16 at 20:37 answer Jun 20 "16 in ~ 20:25 Doug MDoug M 56.2k44 yellow badges2828 silver badges6262 bronze badges \endgroup 5 add a comment | 1 \begingroup A linear transformation is defined by a collection of homogeneous direct (i.e. Every terms have level 1) polynomials, namely of the form$$T(x_1,x_2,x_3)=ax_1+bx_2+cx_3\quad(a,b,c\in\juniorg8.combf R).A and C alone satisfy this criterion.

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BernardBernard
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Is T in A a linear transformation?

Suppose $T:V \rightarrow W$ .Where $V$ and also $W$ space vector spaces over $F$. Let $x_1,x_2,x_3 \in F$ and additionally let $x_4,x_5,x_6 \in F$. So the $(x_1,x_2,x_3) \in V$ and $(x_4,x_5,x_6) \in V$. Now require to inspect that $T((x_1,x_2,x_3)) + T((x_4,x_5,x_6)) = T((x_1+x_4, x_2+x_5,x_3+x_6))$.

We have LHS $= (x_1,0,x_3) + (x_4,0,x_6) = (x_1+x_4,0,x_3+x_6) =$RHS. For this reason this holds by definition of vector addition.

Check linearity because that scalar multiplication:

Let be as above, and suppose $V$ is a vector-space over a field $F$. Climate let $a\in F$. Desire to prove that:

$aT(x_1,x_2,x_3)=T(ax_1,ax_2,ax_3)$. Therefore we have:

LHS $= a(x_1,0,x_3) = (ax_1,0,ax_3) =$RHS. And this holds by vector scalar multiplication and by residential or commercial property of zero in $\juniorg8.combbR$.

Hence this is a linear revolution by definition. In basic you require to display that these 2 properties hold.

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edited Jun 20 "16 in ~ 20:44
answered Jun 20 "16 in ~ 20:34
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