Question

How to check if it's linear?

Answer 1

h(x,y,z)= (x,x+y+z)

Answer 2

You want to show that for any real number \(r\) and any vectors \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) you have \(h(r(x_1,y_1,z_1)+(x_2,y_2,z_2))=rh((x_1,y_1,z_1))+h((x_2,y_2,z_2))\)

Answer 3

\(h(r(x_1,y_1,z_1)+(x_2,y_2,z_2))=h((rx_1+x_2,ry_1+y_2,rz_1+z_2))\\=(rx_1+x_2,rx_1+x_2+ry_1+y_2+rz_1+z_2)\\=(rx_1+x_2,r(x_1+y_1+z_1)+(x_2+y_2+z_2))\\=(rx_1,r(x_1+y_1+z_1))+(x_2,x_2+y_2+z_2)\\=r(x_1,x_1+y_1+z_1)+(x_2,x_2+y_2+z_2)\\=rh((x_1,y_1,z_1))+h((x_2,y_2,z_2))\)

Answer 4

im a bit confused. I learned it differently. Does it matter if it's transposed?

Answer 5

I will try and draw it

Answer 6

what u gotta know is this is a property of princple of linear superposition

Answer 7

if f is a linear fn then
f(x+y) = f(x)+f(y)

Answer 8

correct. so far i made my vector u=(a_1,a_2) and vector v=(b_1,b_2) after adding it i get h(a_1+b_1,a_2,b_2)

Answer 9

yup go on

Answer 10

h(x,y,z)= (x,x+y+z)

Answer 11

I am stuck from there. I do not got a z term.

Answer 12

z=0 for u

Answer 13

how

Answer 14

so h(a_1+b_1,a_1+b_1+a_2+b_2+0)?

Answer 15

???

Answer 16

hi

Answer 17

u still there

Answer 18

h(x,y,z)=(x,x+y+z)
if u and v are vectors
h(u^+v^) =h(<u1,u2,u3>+<v1,v2,v3>)
=h(<u1+v1,u2+v2,u3+v3>)
=(u1+v1,u1+v1+u2+v2+u3+v3)

h(u) + h(v) = (u1,u1+u2 + u3) + (v1,v1+v2+v3)
= (u1+v1,u1+v1+u2+v2+u3+v3)

thus
h(u+v) = h(u) + h(v)

Answer 19

ok how about it h(x,y,z)=h(0,0)

Answer 20

and h = (1,1)

Answer 21

you are going from 3 space to two space. So domain vectors have three components, codomain vectors are 2 dim


you want to show h(a+b) =h(a)+h(b) where a,b are vectors

and if r is a scalar h(ra)=rh(a)


Another way to say this, is, for any vectors a,b and scalar r, we have h(a+rb)=h(a)+rh(b)


note

h((0,0,0)) = (0,0+0+0) = (0,0)

So the zero vector maps to the zero vector.