Exercise I.B.5

Let ϕ,φ,𝜃 be functions given by

ϕ : 2 2, [x y ] [ x x + y ]φ : 3 , [x y z ]zϕ : 2 , [x y ]xy

Show that ϕ and φ are linear transformations, whereas 𝜃 is not.

Answers

1.
Let [x1,x2],[y1,y2] 2 and c be arbitrary. We then have ϕ (c [x1 x2 ] + [y1 y2 ] ) = ϕ ( [cx1 + y1 cx2 + y2 ] ) = [ cx1 + y1 cx1 + y1 + cx2 + y2 ] = c [ x1 x1 + x2 ] + [ y1 y1 + y2 ] = ( [x1 x2 ] )+ϕ ( [y1 y2 ] )

2.
Similarly, we have φ (c [x1 x2 x3 ] + [y1 y2 y3 ] ) = φ ( [cx1 + y1 cx2 + y2 cx3 + y3 ] ) = cx3+y3 = ( [x1 x2 x3 ] )+φ ( [y1 y2 y3 ] )

3.
We have 𝜃 (c [x1 x2 ] + [y1 y2 ] ) = 𝜃 ( [cx1 + y1 cx2 + y2 ] ) = (cx1+y1)(cx2+y2) = c2x 1x2+cx1y2+cx2y1+y1y2

whereas

c𝜃 ( [x1 x2 ] )+𝜃 ( [y1 y2 ] ) = cx1x2+y1y2

Setting x1 := 0 and c,x2,y1,y2 := 1 we get see the inequality of the two quantities.

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2021-10-30 12:11
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