A basic question on isomorphism of two vector spaces Let $V$ and $W$ be of same dimension (over the same field), say dim$V$=dim$W$ = $k < \infty$. Let $T : V \rightarrow W$ be a linear map (which respects vector addition and scalar multiplication). Then if $T$ is $1-1$ then $T$ is onto. How to prove it ? Also, how to prove other way ? i.e. $T$ is onto implies $T$ is $1-1$
Regarding the first part, any $w \in W$ can be written as linear combination of basis vectors of $W$ say $w_i, i=1,\dots,k$, but I can't go to the set $V$ using this $w_i$ because I don't know whether $T$ is onto or not. How to proceed from here ?
 A: This result relies on the following fact. If $T:V\to W$ is a linear transformation between any two vector spaces, then the dimension $\dim(V)=\dim(\ker(T))+\dim(\operatorname{im}(T))$.
You'll also need to know that $T$ is injective iff its kernel is $\{0\}$, and that a subspace $W'$ of $W$ of the same dimension as $W$ is $W$. 
A: Recall that $\dim\ker T+\dim\operatorname{im} T=\dim V$.
In the finite dimensional case we can use  this and the fact that a proper subspace of $W$ has lower dimension. (And in infinite dimensions the claim would indeed not be true)
A: I will prove this without referring to the dimension theorem, just to make it clear:
You need this result: $T(v)=0$ implies $v=0$ iff $T$ is one to one. (This is what is meant by ker$T=\{0\}$). First part: we always have $T(0)=0$. One way to prove it is $T(0)=T(0+0)=T(0)+T(0)$ and subtracting a $T(0)$ both sides give $T(0)=0$. Now if $T$ is one to one $v=0$ can be the only vector such that $T(v)=0$. Second part: Suppose $v=0$ is the only vector such that $T(v)=0$, and suppose $v_1,v_2 \in V$ is such that $T(v_1)=T(v_2)$: then we must have $T(v_1)-T(v_2)=T(v_1-v_2)=0$, and it follows that $v_1-v_2=v=0 \Rightarrow v_1=v_2$ and $T$ is one to one.  
Let $\alpha=\{\alpha_1,\alpha_2,\ldots,\alpha_k\}$ be a basis for $V$. Then we must have $a_1\alpha_1+a_2\alpha_2+\cdots+a_n\alpha_k=0$ iff $a_i=0$ for all $i$ (since the vectors in the basis are linearly independent). 
Now since $T$ is a linear transformation we must have $T(a_1\alpha_1+a_2\alpha_2+\cdots+a_k\alpha_k)=a_1T(\alpha_1)+a_2T(\alpha_2)+\cdots+a_kT(\alpha_k)=T(0)=0$. So then $T(\alpha)=\{T(\alpha_1),T(\alpha_2),\ldots,T(\alpha_k)\}$ must be a linearly independent set, for if there was some $b_1,b_2,\ldots,b_n$ not all zero such that $b_1T(\alpha_1)+b_2T(\alpha_2)+\cdots+b_kT(\alpha_k)=0$ we would have $T(b_1\alpha_1+b_2\alpha_2+\cdots+b_k\alpha_k)=T(0)$ (again by linearity of $T$ and since $T$ is one to one $v=0$ is the only vector such that $T(v)=0$) and so $b_1\alpha_1+b_2\alpha_2+\cdots+b_k\alpha_k=0$ contradicting that $\alpha$ is a basis.
So since $T(\alpha)$ is a linearly independent set with $k$ vectors it is also a basis for $W$ and spans $W$ - that is, $T$ is onto.
Suppose that $T$ is onto. Then for any $w \in W$ there is a $v \in V$ such that $T(v)=w$. Now taking the same basis for $V$ as before we have unique $a_1, a_2, \ldots,a_k$ such that $v=a_1\alpha_1+a_2\alpha_2+\cdots+a_k\alpha_k$. So then again by the linearity of $T$ we have $w=a_1T(\alpha_1)+a_2T(\alpha_2)+\cdots+a_kT(\alpha_k)$. Since $w$ is arbitrary any vector in $W$ can therefore be expressed as a linear combination of $T(\alpha)=\{T(\alpha_1),T(\alpha_2),\ldots,T(\alpha_k)\}$, which means that $T(\alpha)$ spans $W$. Since $T(\alpha)$ is a spanning set consisting of $k$ vectors, $T(\alpha)$ must be a basis for $W$. 
So $w$ is uniquely expressed as $a_1T(\alpha_1)+a_2T(\alpha_2)+\cdots+a_kT(\alpha_k)=T(v)$, and it follows that $T$ is one to one.
A: If you want to do this elementarily, you need just two fundamental results about finitely generated vector spaces (they have counterparts in infinite dimension, but those are less elementary).


*

*Existence of dimension. If a vector space has two finite bases (independent generating sets), these bases have the same number of elements.

*Incomplete basis theorem. If $F$ is an independent set of vectors in $V$, and $G$ is a finite set of vectors generating $V$, then $V$ has a basis $B=F\cup S$ where $S$ is some subset of $G$ (in other words any independent set $F$ can be completed to a basis using elements of $G$).
The second property can be proved from the definitions by formulating a simple algorithm to select$~S$, while the first property requires a bit more work (in which the second property may or may not be instrumental, I forgot); in any case talking about the dimension of a space at all requires knowing the first property.
Now let $T$ be as in the question, and let $B=(b_1,\ldots,b_k)$ be a basis of$~V$. First suppose $T$ is injective (I dislike the term $1-1$). Then $T(b_1),\ldots,T(b_k)$ is a list of independent vectors in$~W$: if some non-trivial linear combination of them would be$~0$, then the corresponding linear combination in$~V$ of the vectors $b_i$ would be annihilated by$~T$, and therefore be$~0$ to begin with by injectiveness of$~T$, but this would contradict the independence of the $b_i$. Now taking some basis $G$ of$~W$, we can by property 2 complete $T(b_1),\ldots,T(b_k)$ by some subset of $G$ to a basis of$~W$. However, by property 1 that basis has to have $k=\dim W$ elements, so the subset of $G$ must be empty. Then $T(b_1),\ldots,T(b_k)$ is already a basis of $~W$, and in particular a generating set of$~W$: every $w\in W$ is a linear combination of them, which is the image by $T$ of the corresponding linear combination of the $b_i$, so $T$ is surjective.
Now conversely suppose that $T$ is surjective, which means that $T(b_1),\ldots,T(b_k)$ is a generating set of$~W$. By property$~2$ we can extend the empty set (which is always independent) to a basis of$~W$ using a subset of $\{T(b_1),\ldots,T(b_k)\}$. By the first property, this subset must have $k=\dim W$ elements, so it can only be the whole set. Then $T(b_1),\ldots,T(b_k)$ are linearly independent, and no non-trivial linear combination of the $b_i$ is annihilated by$~T$. This means that $T$ is injective (if it would map two different linear combinations of the$~b_i$ to the same vector, the difference of those linear combinations would be a non-trivial one annihilated by$~T$).
Comparing with what in the question you reported having tried, it would seem that choosing a basis in$~W$ is not the most fruitful thing to do here. Since $T$ goes from $V$ towards $W$, you can get more mileage out of a basis for$~V$, as you can apply $T$ to that unconditionally. Of course the above argument does use that $W$ admits a basis of $k$ elements (i.e., that $k=\dim W$), but it does not need an explicit example of one.
