# Bases of linearly isomorphic vector spaces

• autre
Since \{Tv_1,\dots,Tv_n\} is linearly dependent, \alpha_{1}T(v_{1})+...+\alpha_{n}T(v_{n})=0 for some nonzero \alpha_{i}. Then there exists a j\in\mathbb{N}, 1\leq j\leq n such that \alpha_{1}T(v_{j}) =-(\alpha_{1}T(v_{1}) +...+\alpha_{j-1}T(v_{j-1})) . Is this on the right track?That could work, but I haven

#### autre

Finite-dimensional $V$ and $W$ are linearly isomorphic vector spaces over a field. Prove that if $\{v_{1},...,v_{n}\}$ is a basis for $V$, $\{T(v_{1}),...,T(v_{n})\}$ is a basis for $W$.

My attempt at a proof:

Let $T:V\rightarrow W$ be an isomorphism and $\{v_{1},...,v_{n}\}$ be a basis for $V$. Since $T$ is an isomorphism, $\forall v\in V,\exists T(v)\in W$. Therefore, if $\{v_{1},...,v_{n}\}$ spans $V$, $\{T(v_{1}),...,T(v_{n})\}$ spans $W$. Since $dim V=dimW$, $\{v_{1},...,v_{n}\}$ is linearly independent, and is a basis for $W$.

I'm pretty sure I went wrong somewhere/claimed something I needed to prove. Any ideas where?

autre said:
Therefore, if $\{v_{1},...,v_{n}\}$ spans $V$, $\{T(v_{1}),...,T(v_{n})\}$ spans $W$.

You just state this, but you never really proved it. In fact, you never used that T is an isomorphism... that given $v\in V$ there exists $T(v)\in W$ is merely a consequence of T being a function. Given $w\in W$, how can you go about determining the coefficients of w in terms of the T(vi)?

Do I use the fact that any $w\in W$ can be written as $w=a_{1}T(v_{1})+...+a_{n}T(v_{n})$ or does that only follow after I show something else?

autre said:
Do I use the fact that any $w\in W$ can be written as $w=a_{1}T(v_{1})+...+a_{n}T(v_{n})$
That's part of what you're trying to prove.

I think you should start by writing down the definition of "basis" that you want to use.

I think you should start by writing down the definition of "basis" that you want to use.

A basis is a linearly independent spanning set. So $\{v_{1},...,v_{n}\}$ spans $V$ and is linearly independent. Does that imply that the set of all $T(v)$ is linearly independent since T is an isomorphism?

autre said:
A basis is a linearly independent spanning set.
Good definition.

autre said:
So $\{v_{1},...,v_{n}\}$ spans $V$ and is linearly independent. Does that imply that the set of all $T(v)$ is linearly independent since T is an isomorphism?
It does, but the question is how. You need to verify that $\{Tv_1,\dots,Tv_n\}$ is a linearly independent set. Can you use the definition of linearly independent (and isomorphism) to do that?

Can you use the definition of linearly independent (and isomorphism) to do that?

Since $T$ is an isomorphism, $\forall v\in V$ there exists only one $T(v)\in W$ such that $v=T(v)$. Since $\{v_{1},...,v_{n}\}$ is linearly independent, $\alpha_{1}v_{1}+...+\alpha_{n}v_{n}=0$ iff $\forall i,\alpha_{i}=0$. Thus $T(\alpha_{1}v_{1}+...+\alpha_{n}v_{n})=T(0)$ and $\{T(v_{1}),...,T(v_{n})\}$ is linearly independent ...am I on the right track?

autre said:
Since $T$ is an isomorphism, $\forall v\in V$ there exists only one $T(v)\in W$ such that $v=T(v)$.
This doesn't quite make sense, but you probably meant something different from what you said. Note that v and Tv aren't even members of the same space. If you meant that for all $v\in V$, there's a unique $w\in W$ such that $Tv=w$, then this is just what it means for T to be a function from V into W.

autre said:
Since $\{v_{1},...,v_{n}\}$ is linearly independent, $\alpha_{1}v_{1}+...+\alpha_{n}v_{n}=0$ iff $\forall i,\alpha_{i}=0$.
This is correct, but I want your first step to be to apply the definition of "linearly independent" directly to $\{Tv_1,\dots,Tv_n\}$ to see if the definition is satisfied.

autre said:
Thus $T(\alpha_{1}v_{1}+...+\alpha_{n}v_{n})=T(0)$
You can't say something like that when you haven't even specified what the v's and a's are. It looks like you have assumed that they are such that $a_1v_1+\cdots+a_nv_n=0$. So you appear to be trying to prove that $\{T(v_{1}),...,T(v_{n})\}$ is linearly independent by having T act only on 0. This can't work.

This is correct, but I want your first step to be to apply the definition of "linearly independent" directly to {Tv1,…,Tvn} to see if the definition is satisfied.

Suppose it is linearly dependent. Then, $\alpha_{1}T(v_{1})+...+\alpha_{n}T(v_{n})=0$ for some nonzero $\alpha_{i}$. Then there exists a $j\in\mathbb{N}, 1\leq j\leq n$ such that $\alpha_{1}T(v_{j}) =-(\alpha_{1}T(v_{1}) +...+\alpha_{j-1}T(v_{j-1}))$. Is this on the right track?

That could work, but I haven't tried to do it that way. This is what I had in mind: Prove that for all $a_1,\dots,a_n\in\mathbb R$ such that $\sum_i a_iTv_i=0$, we have $a_1=a_2=\cdots=a_n=0$. (Replace ℝ with ℂ if it's a vector space over ℂ).

Any idea on how to set that up? I tried to show it by contradiction.

Just start like this: Let $a_1,\dots,a_n$ be arbitrary real numbers such that $\sum_i a_i Tv_i=0$. We have $$0=\sum_i a_i Tv_i=\cdots$$ Now think of something to replace the dots with, and figure out what this means.

it's often easier to prove linear independence rather than "not linear dependence".

make a linear combination of the T(vj) and set it = 0.

T is linear. use this fact to get T "of something".

then use the fact that T is an isomorphism, and thus 1-1,

to show what the "something" must be.

THEN...you should have a statement involving only the vj, and these are linearly independent, so...this should say something about a certain linear combination of them, which will maybe, just maybe, tell us something about the T(vj).