Question

Give an example of a linear transformation T: V -> V which shows that it can happen that T(V) intersect n(T) does not equal zero (is not null).

Answer 1

what math is this? lol.

Answer 2

Haha, linear algebra - theory based. They don't have tutors for the class at OSU. I don't know where to take my questions

Answer 3

oh i see. if you don't get a reply on here then just search the basics on google, it may help you out. i never took this class tbh. or else i wouldve helped you out. :)

Answer 4

Shoot, yeah. I tried google, and like always it rarely helps with this class. The professor wants proofs, and rarely do people ever type up proofs for these random questions. These weekly problem sets are like death. Exactly like death. Haha, but thanks for your help. I'll go and sulk somewhere else

Answer 5

omg i hate proofs! i would literally die ! haha.

Answer 6

lol, are you a math major?

Answer 7

nope. just here to find my answers ! thats what this site is best for.

Answer 8

do u mean: find \(T:V\to V\) such that \(T(V)\cap\ker T\varsupsetneq\{0\}\)?

i mean: that this intersection contains other vector(s) than the zero vector?

Answer 9

@krishnapatel ??

Answer 10

yes! sorry, our book doesn't use (kernel?) but that is exactly what the problem is asking for

Answer 11

give a few minutes.
ok?

Answer 12

sure, thanks!

Answer 13

r u still there?

Answer 14

sorry. it took a bit long for me to figure this out.
do u know derivatives?

Answer 15

no its fine, yes i know of derivatives and also that derivatives are a linear transformation

Answer 16

great

Answer 17

does the space V need be finite-dimensional?

Answer 18

From my understanding, yes. But I may be wrong and the question might be looking for such an answer that requires V to be of infinite dimension. A relevant theorem (called the Fundamental Theorem( states: dim T(V) + dim n(T) = dim V. I was thinking we might have to use this somewhere

Answer 19

there is no need for this theorem

Answer 20

well let's consider \(V=\{p(x)\in \mathbb{R}[x]:\deg p(x)\leq n\}\) if V has to be finite-dimensional.
if V has to be infinite-dimensional u just put \(V=\mathbb{R}[x]\).
then let \(D:V\to V\) be defined by \(D(p(x))=p'(x)\)

Answer 21

got it?

Answer 22

I understand the transformation, but wouldn't the transformation be from a n dimensional space into a n-1 dimensional space? (Don't you lose a dimension when you take the derivative of a polynomial?)

Answer 23

i a way, u r right. but the problem doesn't say the tranformation has to be onto, i.e. T(V)=V.

Answer 24

Hmm, but then what does D: V ->V represent?

Answer 25

instead of say V -> W.

Answer 26

just a linear transformation with domain and codomain equal

Answer 27

Ah, okay that makes sense. But then what is T(V) or n(T)? would n(T) be a P(n) where n = 0 (Polynomial with degree zero = a constant)?

Answer 28

ok. so
\[ \large \ker D=\{p(x)=\alpha:\alpha\in\mathbb{R}\} \]
i.e., the kernel is composed by the constant polynomials.
ok?

Answer 29

yes!

Answer 30

great. now the polynomials of the form \(p(x)=kx\) have image
\[ \large D(p(x))=D(kx)=(kx)'=k \]
a constant polynomial.
ok?

Answer 31

yep!

Answer 32

so we did it!

Answer 33

for this linear transformation
\[ \large D(V)\cap\ker D=\ker D\varsupsetneq0 \]

Answer 34

Sorry that I'm taking a little longer to grasp this. I thought we said that D(V) consists of polynomials of form kx, where as ker D consists of polynomials of form k(x^0). then how is the intersection of those two polynomials ker D?

Answer 35

Ohhh, is it for all x such that x = 1?

Answer 36

no. \(x\) is an indetermminate. so it is FOR ALL x.

Answer 37

Yes, actually I think I understand. Correct me if I'm wrong. We have ker D to be a subset of D(V) where the intersection of ker D and D(V) is the set ker D itself.

Answer 38

which consists of the polynomial functions of the form k(x^0)

Answer 39

yes

Answer 40

wow that's beautiful. thank you so much for your help!

Answer 41

u r welcome