Question

Linear transformations : Show that...
see attachment

Answer 1

Could you recall what linearly independent means?

Answer 2

Linear independence is just such that any one vector in the set, isn't a linear combination of the other vectors? I still do not know how to prove, that after a linear transformation, all vectors are still independent

Answer 3

Well then, let's see, shall we?
We know that v1, v2, v3... vr are linearly independent, right?

Answer 4

Still there? You need to pay attention to the details here...

Answer 5

Yes

Answer 6

Okay, so let's check their respective images under the linear transformation...
Consider
T(v1), T(v2), T(v3), ... T(vr)
catch me so far?

Answer 7

Yes

Answer 8

Okay, so, let's proceed by contradiction :)
Suppose there is a linear combination of these vectors that produce the zero vector:
a1T(v1) + a2T(v2) + ... + arT(vr) = 0

still with me?

Answer 9

okay

Answer 10

Well, what do we know about the linear transformations (such as T)

We know that constants can move in and out of the transformation:

aT(v) = T(av) right?

Answer 11

Yep

Answer 12

Okay, so this equation
a1T(v1) + a2T(v2) + ... + arT(vr) = 0
May be written as:

T(a1 v1) + T(a2 v2) + ... + T(ar vr) = 0

yes?

Answer 13

Yes

Answer 14

Well, maybe you can proceed from here?
Perhaps use the other property of linear transformations...

Answer 15

@dw3
well?

Answer 16

I don't think I understand

Answer 17

Okay, so another property of linear transformations (which you forgot -_-)
is this:

T(u) + T(v) = T(u+v)

Answer 18

T(a1v1+⋯+arvr)=0

But what does this mean :/

Answer 19

Well, in a linear transformation, ONLY the zero vector could be mapped to the zero vector, right?

Answer 20

Here, let me prove it.
Suppose we want to find T(0)

T(0) = T(v - v)
= T(v) - T(v)
= 0

Answer 21

Ah! i get it

Answer 22

So, This implies that
a1v1 + a2v2 + ... arvr = 0
And this is...?

Answer 23

linearly dependent?

Answer 24

a contradiction, silly :>

Didn't we just assume that v1, v2, ...vr are linearly independent?

Answer 25

oh, >.<
Thanks a lot!

Answer 26

Wait a minute, there might be a slight problem....

Answer 27

We have to assume that the transformation T is at least injective, otherwise, this won't work....