Do Linear Operators Equate on All Vectors if They Match on a Basis Set?

[pyroknife]

Suppose that T1: V → V and T2: V → V are
linear operators and {v1, . . . , vn} is a basis for V .
If T1(vi) = T2(vi ), for each i = 1, 2, . . . , n, show
that T1(v) = T2(v) for all v in V .


I don't understand this question.
They said If T1(vi) = T2(vi ), for each i = 1, 2, . . . , n
wouldn't that mean T1(v)=T2(v) already? I don't get what I have to prove here.


Isn't this just saying
T1(v1)=T2(v1)
T1(v2)=T2(v2)
.
.
.
T1(vn)=T2(vn)?

 

[LCKurtz]

Suppose that T1: V → V and T2: V → V are
linear operators and {v1, . . . , vn} is a basis for V .
If T1(vi) = T2(vi ), for each i = 1, 2, . . . , n, show
that T1(v) = T2(v) for all v in V .


I don't understand this question.
They said If T1(vi) = T2(vi ), for each i = 1, 2, . . . , n
wouldn't that mean T1(v)=T2(v) already? I don't get what I have to prove here.


Isn't this just saying
T1(v1)=T2(v1)
T1(v2)=T2(v2)
.
.
.
T1(vn)=T2(vn)?

No. You are given that T1 and T2 give the same thing for the basis vectors. What about any other vector v? Show T1(v) = T2(v).

 

[pyroknife]

T1(v) = T1(c1v1 + ... + cnvn)
= T1(c1v1) + ... + T1(cnvn)
= c1T1(v1) + ... + cnT1(vn)
= c1T2(v1) + ... + cnT2(vn)
= T2(c1v1) + ... + T2(cnvn)
= T2(c1v1 + ... + cnvn)
= T2(v)

 

[LCKurtz]

That's it.

 

[Fredrik]

T1(v) = T1(c1v1 + ... + cnvn)
= T1(c1v1) + ... + T1(cnvn)
= c1T1(v1) + ... + cnT1(vn)
= c1T2(v1) + ... + cnT2(vn)
= T2(c1v1) + ... + T2(cnvn)
= T2(c1v1 + ... + cnvn)
= T2(v)

Looks good. The only thing I would add is a statement like "Let v be an arbitrary member of V" or "For all v in V, we have..."

Edit: D'oh...too slow.