Linear Operators and Dependence

[charlies1902]

Suppose that T: V →V is a linear operator and
{v1, . . . , vn} is linearly dependent. Show that
{T (v1), . . . , T (vn)} is linearly dependent.



I'm pretty lost as to how to even go about doing this problem, but I'll take a crack at it.

I'm not sure what the operator "T:V→V" means. It seems it's transforming V to V, so I don't get the point.
If that's what it means then:
{T (v1), . . . , T (vn)} = {v1, . . . , vn}
Thus {T (v1), . . . , T (vn)} is linearly dependent?

 

[LCKurtz]

Suppose that T: V →V is a linear operator and
{v1, . . . , vn} is linearly dependent. Show that
{T (v1), . . . , T (vn)} is linearly dependent.
I'm pretty lost as to how to even go about doing this problem, but I'll take a crack at it.

I'm not sure what the operator "T:V→V" means

T is given to be a linear operator from V to V. I have a couple of things for you:
1. Show us the definition of a linear operator.
2. Show us the definition that {v1, . . . , vn} is linearly dependent.
Once you have those, write down what you are given and what you have to prove. Then let's see what happens.

 

[charlies1902]

1. This is copied from the book but I understand it "Let V and W be vector spaces. The mapping T: V → W
is called a linear transformation if and only if
T (cu + v) = cT (u) + T (v)
for every choice of u and v in V and scalars c in
. In the case for which V = W,
then T is called a linear operator."

2. There are vector(s) in {v1, . . . , vn} such that one is the linear combination of another.

if we write c1v1+...+cnv2=0, where c1...cn scalars. If they're linearly dependent they those scalars are not unique meaning that linear combos exist..
This is what my friend said:
{v1, . . . , vn} = {v1, ..., va, c*va, ..., vn} where va is a vector and there's a multiple of it somewhere
so T:{v1, . . . , vn} gives {T(v1), ..., T(vn)} = {T(v1), ..., T(va), T(c*va), ..., T(vn)} = {T(v1), ..., T(va), c*T(va), ..., T(vn)}.
c*T(va) is still a multiple of T(va), thus it is still linearly dependent.

 

[LCKurtz]

1. This is copied from the book but I understand it "Let V and W be vector spaces. The mapping T: V → W
is called a linear transformation if and only if
T (cu + v) = cT (u) + T (v)
for every choice of u and v in V and scalars c in . In the case for which V = W,
then T is called a linear operator."

This is usually written as the two conditions: T(cv) = cT(v) and T(u+v) = T(u)+T(v). Alternatively it is sometimes written as the single condition T(cu + dv) = cT(u)+dT(v). That makes T linear and it doesn't matter whether V=W or not.

2. There are vector(s) in {v1, . . . , vn} such that one is the linear combination of another.

if we write c1v1+...+cnv2=0, where c1...cn scalars. If they're linearly dependent they those scalars are not unique meaning that linear combos exist..

Those are consequences of linear dependence. The actual definition is that there exist constants $c_1,c_2,...,c_n$ not all zero such that $c_1v_2+c_2v_2+...,c_nv_n=\theta$, the zero vector.

Now, using the definitions I have written, write down what you are given and what you have to prove. Then see if you can prove it using the definitions.

 

[charlies1902]

This is usually written as the two conditions: T(cv) = cT(v) and T(u+v) = T(u)+T(v). Alternatively it is sometimes written as the single condition T(cu + dv) = cT(u)+dT(v). That makes T linear and it doesn't matter whether V=W or not.

Those are consequences of linear dependence. The actual definition is that there exist constants $c_1,c_2,...,c_n$ not all zero such that $c_1v_2+c_2v_2+...,c_nv_n=\theta$, the zero vector.

Now, using the definitions I have written, write down what you are given and what you have to prove. Then see if you can prove it using the definitions.

If there exist constants c1,c2,...,cn not all zero such that c1v2+c2v2+...,cnvn=θ, the zero vector, then that would mean some vector is a linear combination of another.

Wouldn't this proof be valid?
{v1, . . . , vn} = {v1, ..., va, c*va, ..., vn} where va is a vector and there's a multiple of it somewhere
so T:{v1, . . . , vn} gives {T(v1), ..., T(vn)} = {T(v1), ..., T(va), T(c*va), ..., T(vn)} = {T(v1), ..., T(va), c*T(va), ..., T(vn)}.
c*T(va) is still a multiple of T(va), thus it is still linearly dependent.

 

[LCKurtz]

Those are consequences of linear dependence. The actual definition is that there exist constants $c_1,c_2,...,c_n$ not all zero such that $c_1v_2+c_2v_2+...,c_nv_n=\theta$, the zero vector.

Now, using the definitions I have written, write down what you are given and what you have to prove. Then see if you can prove it using the definitions.

If there exist constants c1,c2,...,cn not all zero such that c1v2+c2v2+...,cnvn=θ

Good. That is what you are given.

the zero vector, then that would mean some vector is a linear combination of another.

Yes, it implies that, but let's use the definition.

Wouldn't this proof be valid?

There is the germ of an idea in that argument but it certainly isn't adequate. The problem is that you wrote down what you were given but you haven't written down what you are to prove, using the definitions. Write down, using the definition, what you are to prove.

 

[charlies1902]

Suppose that T: V →V is a linear operator and
{v1, . . . , vn} is linearly dependent. Show that
{T (v1), . . . , T (vn)} is linearly dependent.

I must show that for {T (v1), . . . , T (vn)} there exist constants b1, b2, ..., bn not all zero such that b1T(v1)+b2T(V2),...,bnT(Vn)=0

 

[LCKurtz]

Suppose that T: V →V is a linear operator and
{v1, . . . , vn} is linearly dependent. Show that
{T (v1), . . . , T (vn)} is linearly dependent.

I must show that for {T (v1), . . . , T (vn)} there exist constants b1, b2, ..., bn not all zero such that b1T(v1)+b2T(V2),...,bnT(Vn)=0

Good. Now start with what you are given, apply T to it, use the properties of a linear function step by step and see if you can show that.

 

[HallsofIvy]

Suppose that T: V →V is a linear operator and
{v1, . . . , vn} is linearly dependent. Show that
{T (v1), . . . , T (vn)} is linearly dependent.

I must show that for {T (v1), . . . , T (vn)} there exist constants b1, b2, ..., bn not all zero such that b1T(v1)+b2T(V2),...,bnT(Vn)=0

Yes, and you are given that the {v1,..., vn} are linearly dependent- that is,
$$b1v1+ b2v2+ ...+ bnvn= 0$$
for some b1, b2, ..., bn that are not all 0. What do you get if you apply T to both sides of that?

 

[charlies1902]

Good. Now start with what you are given, apply T to it, use the properties of a linear function step by step and see if you can show that.

Yes, and you are given that the {v1,..., vn} are linearly dependent- that is,
$$b1v1+ b2v2+ ...+ bnvn= 0$$
for some b1, b2, ..., bn that are not all 0. What do you get if you apply T to both sides of that?

T(b1v1)+T(b2v2)+...+T(bnvn)=T(0)
b1T(V1)+b2T(v2)+...bnT(vn)=0
= b1V2+b2V2+...bnVn=0,
thus it is still linearly dependent.Sweet, I think that's correct. Thanks for all the help.

 

[LCKurtz]

T(b1v1)+T(b2v2)+...+T(bnvn)=T(0)
b1T(V1)+b2T(v2)+...bnT(vn)=0
= [STRIKE]b1V2+b2V2+...bnVn=0,[/STRIKE]
thus the set {T(v1),...,T(vn)} is [STRIKE]still[/STRIKE] linearly dependent.


Sweet, I think that's correct. Thanks for all the help.

I edited it a bit. You're welcome.

 

[HallsofIvy]

T(b1v1)+T(b2v2)+...+T(bnvn)=T(0)
b1T(V1)+b2T(v2)+...bnT(vn)=0

Yes that's what you want.

I don't know why you added this next equation which just restates what you are given.

= b1V2+b2V2+...bnVn=0,
thus it is still linearly dependent.


Sweet, I think that's correct. Thanks for all the help.

Note that the other way is not necessarily correct: if v1, v2, ..., vn are independent it does not follow that Tv1, Tv2, ..., Tvn are independent.

 

[charlies1902]

I edited it a bit. You're welcome.

Thanks again.

Yes that's what you want.

I don't know why you added this next equation which just restates what you are given.

Note that the other way is not necessarily correct: if v1, v2, ..., vn are independent it does not follow that Tv1, Tv2, ..., Tvn are independent.

I saw this as a definition in my textbook, but they didn't prove it.

If v1, v2, ... vn are independent then
c1v1+c2v2+...+cnv2=0, where c1, c2, ...cn are scalars and =0..

T(c1v1)+T(c2v2)+...+T(cnvn)=T(0)
c1T(v1)+c2T(v2)+...+cnT(vn)=0

For the case of T: V →V, wouldn't they still be linearly independent?

 

[Mark44]

Thanks again.


I saw this as a definition in my textbook, but they didn't prove it.

Definitions are never proved.

If v1, v2, ... vn are independent then
c1v1+c2v2+...+cnv2=0, where c1, c2, ...cn are scalars and =0..

T(c1v1)+T(c2v2)+...+T(cnvn)=T(0)
c1T(v1)+c2T(v2)+...+cnT(vn)=0

For the case of T: V →V, wouldn't they still be linearly independent?

 

[HallsofIvy]

Thanks again.


I saw this as a definition in my textbook, but they didn't prove it.

If v1, v2, ... vn are independent then
c1v1+c2v2+...+cnv2=0, where c1, c2, ...cn are scalars and =0..

T(c1v1)+T(c2v2)+...+T(cnvn)=T(0)
c1T(v1)+c2T(v2)+...+cnT(vn)=0

For the case of T: V →V, wouldn't they still be linearly independent?

If and only if T is invertible.

For example, Tv= 0 for all v is a linear transformation but it maps a set of independent vectors {v1, v2, ..., vn} to T(v1)= 0, T(v2)= 0, ..., T(vn)= 0 so certainly NOT a set of independent vectors.