# Thread: Linear Transformation - Questions reg Chr and Minimal Polynomial

1. ## Linear Transformation - Few quick questions

Let V - finite dimensional vector space over field, F
T,S be linear transformation from V into/onto V
I be the identity tranformation

Q1: m(x) is minimal polynomial of T
p(x) is characteristic polynomial of T
if q(x) is an irreducible polynomial over F and q(x)|p(x), prove that q(x)|m(x)
as well.

Q2: If ST=I => TS = I
(basically I want to prove if a transformation is right or left invertible, it is invertible)

Q3: T^n = 0 => T=0. I am not sure of this one, but couldn't find the reason/counter-example

Any pointers would be welcome. I am just picking up the subject. So these might really be elementary question but will appreciate any help here.

Thanks

2. Originally Posted by aman_cc
Q2: If ST=I => TS = I
(basically I want to prove if a transformation is right or left invertible, it is invertible)
Please validate if this argument is correct.
ST = I => T(S(v)) = v for all v in V
Thus T is onto and hence invertible. Let inverse of T be T'

so TS = TS(TT') = T(ST)T' = TIT' = I (Hence Proved)

Looks fine to me. Any better argument?

3. Originally Posted by aman_cc
Let V - finite dimensional vector space over field, F
T,S be linear transformation from V into/onto V
I be the identity tranformation

Q1: m(x) is minimal polynomial of T
p(x) is characteristic polynomial of T
if q(x) is an irreducible polynomial over F and q(x)|p(x), prove that q(x)|m(x)
as well.
Hint: in general if $\dim_F V = n,$ then $p(x) \mid (m(x))^n.$

Q2: If ST=I => TS = I
(basically I want to prove if a transformation is right or left invertible, it is invertible)
$S$ is 1-1 because it's onto. we also have $STS(v)=S(v),$ for all $v \in V,$ because $ST=\text{id}.$ thus $TS(v)=v$ because $S$ is 1-1.

Q3: T^n = 0 => T=0. I am not sure of this one, but couldn't find the reason/counter-example

Any pointers would be welcome. I am just picking up the subject. So these might really be elementary question but will appreciate any help here.

Thanks
you forgot to mention what $n$ is! whatever it is, the implication is false. for example, on $V=\mathbb{R}^2,$ define $T(x,y)=(y,0).$ then $T^2=0$ but $T \neq 0.$

4. Thanks very much. I'll try your hint to Q1.

Let T:V->V be a linear transformation. Am I correct in saying:
T is onto IFF T is one-one

Proof: Let v1,v2, v3...,vn be basis of V
T is onto => T(v1), T(v2), ... T(vn) span V hence is a basis of V. So if T(v) = 0 => v = 0. Which further => T is one-one

T is one-one. T(v1), T(v2)....T(vn) is basis for range of T which is a equal to/subspace of V. But any subspace of V with n basis = V itself hence T is onto.

Is my reasoning correct?

5. Originally Posted by aman_cc
Thanks very much. I'll try your hint to Q1.

Let T:V->V be a linear transformation. Am I correct in saying:
T is onto IFF T is one-one

Proof: Let v1,v2, v3...,vn be basis of V
T is onto => T(v1), T(v2), ... T(vn) span V hence is a basis of V. So if T(v) = 0 => v = 0. Which further => T is one-one

T is one-one. T(v1), T(v2)....T(vn) is basis for range of T which is a equal to/subspace of V. But any subspace of V with n basis = V itself hence T is onto.

Is my reasoning correct?
it's correct. you may also use the rank-nullity theorem for a faster proof.