Question

determine whether the following function is a linear transformation,justify your answer.
T:V>R,where V is an inner product space and T(U)=//U//

Answer 1

the transformation is\[T(\vec u)=\|\vec u\|\]?

Answer 2

yep

Answer 3

you know the two tests we need to do to see if this is a linear transformation?

Answer 4

yes i do

Answer 5

\[T(\vec u+\vec v)=T(\vec u)+T(\vec v)\]and\[T(c\vec u)=cT(\vec u)\]

Answer 6

so let's try the second one first; seems easier....

Answer 7

ok

Answer 8

\[\vec u=\langle a_1,a_2,...a_n\rangle\]\[c\vec u=\langle ca_1,ca_2,...ca_n\rangle\]\[T(c\vec u)=\sqrt{c^2a_1^2+c^2a_2^2+...+c^2a_n^2}\]you tell me, is that the same as\[cT((\vec u)\]?

Answer 9

absolutely the same

Answer 10

I agree, so on to the other test...

Answer 11

\[\vec u+\vec v=\langle a_1+b_1,a_2+b_2,...,a_n+b_n\rangle\]\[T(\vec u+\vec v)=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2+...+(a_n+b_n)^2}\]is that equal to\[\sqrt{a_1^2+a_2^2+...+a_n^2}+\sqrt{b_1^2+b_2^2+...+b_n^2}\]?

Answer 12

in nt sure

Answer 13

\[T(\vec u+\vec v)=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2+...+(a_n+b_n)^2}\]\[=\sqrt{a_1^2+2a_1b_1+b_1^2+a_2^2+2a_2b_2+b_2^2+...+a_n^2+2a_nb_n+b_n^2}\]looks to me like we sonw be able to combine the radicals in\[\sqrt{a_1^2+a_2^2+...+a_n^2}+\sqrt{b_1^2+b_2^2+...+b_n^2}\]or get rid of those cross terms \(2a_kb_k\), so I would say they seem not to be the same

Answer 14

use u= -v

Answer 15

oh yeah that would have made it way more obvious :P

Answer 16

so it doesnt satisfy the two conditions?

Answer 17

try phi's suggestion and it should be clear that it does not satisfy the last condition

Answer 18

\[\vec v=-\vec u\]\[T(\vec u+(-\vec u))=T(0)=0\neq\|\vec u\|+\|-\vec u\|=2\|\vec u\|\]hence\[T(\vec u+\vec v)\neq T(\vec u)+T(\vec v)\]

Answer 19

meaning is not a linear transformation?

Answer 20

if it fail either test it is not a linear transformation

Answer 21

...so yeah