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 ?


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 $W$.

  • $\begingroup$ have not understood anything. Are you talking about the first part ? Can't it be just proved using the definition of linear map and $T(x) = \theta_w$ implies $x=\theta_v$ (i.e. injective). Can you plz be more elaborate ? $\endgroup$ – RIchard Williams Aug 15 '13 at 6:32
  • $\begingroup$ it can't just be proven without using any special property of finite numbers, as the result is not true for infinite dimensional spaces. You will have to first come to grips with the first result I quote. Preferably look in any textbook and find a proof of it. Then you can think about how to use it, and the other results I mention, to prove what you need. $\endgroup$ – Ittay Weiss Aug 15 '13 at 6:37
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    $\begingroup$ @IttayWeiss typo in $\dim V$ equality. $\endgroup$ – pppqqq Aug 15 '13 at 19:09
  • $\begingroup$ @pppqqq thanks! $\endgroup$ – Ittay Weiss Aug 15 '13 at 21:37

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)


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.

  • $\begingroup$ I think this answer says essentially the same as the one I gave (quite a bit later), so I'll give it +1. However "since $T$ is $1-1$, each $T(\alpha_i)$ is a unique vector in $W$" is only confusing: applying a map to an element always gives a unique image (which is not the same as being the image of a unique element). $\endgroup$ – Marc van Leeuwen Aug 16 '13 at 10:00
  • $\begingroup$ Yes thanks Marc - I edited that out - the 1-1 really comes in where the only solution to $T(v)=0$ is $v=0$...thanks for pointing that out... $\endgroup$ – Christiaan Hattingh Aug 16 '13 at 12:02

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).

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

  2. 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.


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