Prove Basis to Basis Isomorphism

[iamalexalright]

Homework Statement

Let t \in L(V,W). Prove that t is an isomorphism iff it carries a basis for V to a basis for W.

Homework Equations

L(V,W) is the set of all linear transformations from V to W

The Attempt at a Solution

So I figured I would assume I have a transformation from a basis for V to a basis for W and prove that it is bijective (and hence isomorphic).

So let B_{v} = \{v_{i} | i \in I\} be a basis for V and
B_{w} = \{w_{j} | j \in J\} be a basis for W

So t is surjective if im(t) = W so...
im(t) = \{tv_{i} | v_{i} \in B_{v}\} =
\{w_{j} | w_{j} \in B_{w}\}
Since B_w spans W can I simply say then that, given above, the next line would be:
= W
so im(t) = W

Now to prove it is injective I need to show that ker(t) = {0}
tv_{i} = tv_{j} implies
t(v_{i} - v_{j}) = 0 implies
v_{i} - v_{j} \in ker(t)

If ker(t) = {0} then implies v_i = v_j but this doesn't seem like a formal way to prove this... any help?

 

[Robert1986]

Homework Statement

Let t \in L(V,W). Prove that t is an isomorphism iff it carries a basis for V to a basis for W.

Homework Equations

L(V,W) is the set of all linear transformations from V to W

The Attempt at a Solution

So I figured I would assume I have a transformation from a basis for V to a basis for W and prove that it is bijective (and hence isomorphic).

So let B_{v} = \{v_{i} | i \in I\} be a basis for V and
B_{w} = \{w_{j} | j \in J\} be a basis for W

So t is surjective if im(t) = W so...
im(t) = \{tv_{i} | v_{i} \in B_{v}\} =
\{w_{j} | w_{j} \in B_{w}\}
Since B_w spans W can I simply say then that, given above, the next line would be:
= W
so im(t) = W

Now to prove it is injective I need to show that ker(t) = {0}
tv_{i} = tv_{j} implies
t(v_{i} - v_{j}) = 0 implies
v_{i} - v_{j} \in ker(t)

If ker(t) = {0} then implies v_i = v_j but this doesn't seem like a formal way to prove this... any help?

So, you have assumed that if v_i is a basis element of V then its image under t is w_i, which is a basis element of W. Now, observe that, by assumption, no two basis elements of V are mapped to the same basis element in W. Now, write your v_i,v_j as a linear combo of vectors in V and then use the linearity of t to distribute it across the linear combo.

Also, you need to prove that if t is an isomorphism, then it maps a basis of V to a basis of W, that is, the image of the basis of V under T forms a basis of W.

 

[iamalexalright]

So i did assume it was 1-1.

okay, well for v_i in the Basis of V:

t(a_{1}v_{1} + ... + a_{n}v_{n} =
a_{1}tv_{1} + ... + a_{n}tv_{n}

If it is an isomorphism then the ker(t) should be zero:

ker(t) = \{v \in B_{v} | tv = 0\}
so:
a_{1}tv_{1} + ... + a_{n}tv_{n} = 0
since I assume t maps to a basis in W then the only solution to this is when a1...an = 0 so that implies that the ker(t) = {0}, yes?

Now the image of t is simply tv, and since I assume tv is in a basis of W then the image spans W so I can write im(t) = W. Correct?

I still feel i am wrong

 

[Robert1986]

So i did assume it was 1-1.

okay, well for v_i in the Basis of V:

t(a_{1}v_{1} + ... + a_{n}v_{n} =
a_{1}tv_{1} + ... + a_{n}tv_{n}

If it is an isomorphism then the ker(t) should be zero:

ker(t) = \{v \in B_{v} | tv = 0\}
so:
a_{1}tv_{1} + ... + a_{n}tv_{n} = 0
since I assume t maps to a basis in W then the only solution to this is when a1...an = 0 so that implies that the ker(t) = {0}, yes?

Now the image of t is simply tv, and since I assume tv is in a basis of W then the image spans W so I can write im(t) = W. Correct?

I still feel i am wrong

You have done enough to show that t is 1-1 and onto and hence an isomorphism. The only problem is that you need to be a *little* more formal in your proof that it is onto, but you have the idea.

I'd start by writing every element in W as the linear combo of a basis for W. Which basis? Well, given your assumptions (that is, T takes a basis for V to a basis for W) which basis do you think you should use?

 

[iamalexalright]

Yeah, when I write it up I will take care of being a bit more formal

let u \in W
then
u = a_{1}w_{1} + ... + a_{n}w_{n}
where w is in a basis for W
then since w_{i} = b_{1}tv_{1} + ... + b_{n}tv_{n} (v in basis for V)

we can combine and get:
u = a_{1}[b_{11}tv_{1} + ... b_{n1}tv_{n}] + ... + a_{n}[b_{1n}tv_{1} + ... + b_{nn}tv_{n}]

simplifying:
u = c_{1}tv_{1} + ... + c_{n}tv_{n} =

Thanks for the help

 

[Robert1986]

One simplifying step:

You are saying that T maps SOME basis of V to SOME basis of W. So, you don't know what the bases are, but you know there exists a basis for V and a basis for W such that T(v_i)=w_i. Now, write your arbitrary element of W as a linear combination of the w_i's. This way you can skip the portion about combining the linear combinations.Now, if you haven't, you need to prove the other way of this bi-conditional statement. That is, assume that it is an isomorphism and show that it maps a basis of V onto a basis of W.

 

[iamalexalright]

So if it is an isomorphism we know that

im(t) = W so each w in W can be written as a combination a_1*tv_1 + ... + a_n*tv_n

since the kernel(t) = {0} this implies only the zero vector is a solution to:
a_1* tv_1 + ... a_n* tv_n = 0

so then these vectors v_n are linearly independent. I don't know how to show that these span V however.

Eh... not so sure about these half of it..

 

[Robert1986]

So if it is an isomorphism we know that

im(t) = W so each w in W can be written as a combination a_1*tv_1 + ... + a_n*tv_n

since the kernel(t) = {0} this implies only the zero vector is a solution to:
a_1* tv_1 + ... a_n* tv_n = 0

so then these vectors v_n are linearly independent. I don't know how to show that these span V however.

Eh... not so sure about these half of it..

It hasn't been said explicitly, but if T is an isomorphism from V to W, then what can you say about the dimensions of V and W. That is, which one, if any, is true:
dim(V) < dim(W)
dim(V) >dim(W)
dim(V) <= dim(W)
dim(V) >=dim(W)
dim(V) = dim(W)
(HINT: use nullity plus rank theorem and use the fact that ker(T)={0})

Now, as you pointed out a_1* tv_1 + ... a_n* tv_n = 0 implies that a_1=...=1_n=0. What does this say about these vectors? Now, look to see how many of these vectors you have. Do you have a basis?

 

[iamalexalright]

The dimensions are the same

if the coeffecients being zero is the only way the linear combination of vectors is zero then the vectors are linearly independent. I have n vectors so assuming the dimension of my space is n then I have a basis.

 

[Robert1986]

The dimensions are the same

if the coeffecients being zero is the only way the linear combination of vectors is zero then the vectors are linearly independent. I have n vectors so assuming the dimension of my space is n then I have a basis.

Perfect! You have shown a)T is 1-1 b)every w in W is a linear combo of the T(v_i)'s, so T is onto.

You're done!

 

[iamalexalright]

: D
Thanks so much for the help!