# existence of linear transformation

This is a simple question in linear algebra but I just had a hard time thinking about the logic.

Let V and W be two vector spaces over a field F. For simplicity, we assume they have the same finite dimensions and any linear transformation $$T:V \rightarrow W$$, null space is {$$0$$}. Let {$$v_1$$, $$v_2$$, ..., $$v_n$$} be a basis for V. Then by theorem, any linear transformation $$T:V \rightarrow W$$, {$$T(v_1)$$, $$T(v_2)$$, ..., $$T(v_n)$$} should be a basis for W. However, theorem of existence of linear transformation states that there exists precisely one linear transformation
$$T:V \rightarrow W$$ such that $$T(v_i) = w_i$$ where $$w_i, i = 1,..,n$$ is any vectors in W. If {$$w_1$$, $$w_2$$, ..., $$w_n$$} are linear dependent, then one of two theorms is wrong. Did I omit something?

edit: the first theorem can be derived from $$N(T)$$ = {$$0$$}, then T is injective, then T is an isomorphism given dim(V) = dim(V) < $$\infty$$.

• It is not true that if $V$ and $W$ have the same dimension then any linear map $T\colon V\to W$ has $\{0\}$ null space. – egreg Feb 14 at 22:34
• @egreg Those are the assumption. – YellowRiver Feb 14 at 22:46
• You read them wrongly, I'm afraid. The only vector space $V$ such that every linear map $T\colon V\to V$ is injective is the trivial space. – egreg Feb 14 at 22:52
• @egreg I mean that "$V$ and $W$ having the same dimension" and "map $T: V \to W$ having {$0$} null space" are assumptions. – YellowRiver Feb 14 at 22:57
• I think the question really is that there doesn't exist a linear transformation that maps independent basis to dependent vectors when it is an isomorphism. – YellowRiver Feb 14 at 23:03

The image vectors $$\{T(v_{1}),\ldots, T(v_{n})\}$$ can form a basis only if the kernel of $$T$$ is zero. This is not true anymore if $$\{w_{1},\ldots,w_{n}\}$$ are linearly dependent, because by definition you can find scalars $$\lambda_{i}$$'s not all zero such that $$\lambda_{1}w_{1}+\cdots+\lambda_{n}w_{n}=0$$ Then $$T(\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n})=0$$, so we have a non-zero element in the kernel (if $$\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n}=0$$ then all $$\lambda_{i}$$'s would be zero by linear independence of the $$v_{i}$$'s).

So in the first theorem you need to assume that $$T$$ is an isomorphism (or equivalently that it has trivial kernel if both spaces have the same dimension).

• I believe I just stated in that way: "null space is {$0$}"/trivial kernel and both spaces have the same dimensions. – YellowRiver Feb 14 at 22:07
• Yes, after your last edit the statement is more clear. So you see that both theorems are fine, but the point is that if the $w_{i}$ are linearly dependent, then the corresponding $T$ does have some non-trivial kernel, so you cannot apply the first theorem – Pedro Feb 14 at 22:12

There's a few things to address here. First, the null space is defined in terms of an operator, and is not an intrinsic property of spaces themselves.

Secondly I am not sure where you are getting this theorem, but I believe you may be misinterpreting it. If $$T:V\to W$$ is an isomorphism then yes, $$\{Tv_i\}_{i=1}^n$$ is a basis for for $$W$$, but if not then no. Just consider the $$0$$ operator.

Thirdly if $$\{w_i\}_{i=1}^n$$ is a basis then they will not be linearly dependent.

• I agree with all your points... But my setting is exactly $T:V\to W$ is an isomorphism, $\{Tv_i\}_{i=1}^n$ is a basis for for $W$ and $\{w_i\}_{i=1}^n$ is not a basis. – YellowRiver Feb 14 at 22:10
• @YellowRiver For a given set $\{w_i\}_{i=1}^n$ yes then there is a unique operator $S:V\to W$ satisfying $Sv_i=w_i$, but there is no reason why this operator has to equal $T$? In fact for your given $T$ we know that this can't be $S$ precisely because it maps a linearly independent set to a linearly independent set. – K.Power Feb 14 at 22:16
• Nice! I just realized the importance of your first point. When setting a null space to be {$0$}, choice of operator T is limited. – YellowRiver Feb 14 at 22:25

Suppose that the linear transformation being considered is one that takes every vector in $$V$$ to the $$0$$ of $$W$$. Clearly, it could never result in a basis. The theorem that all linear transformations map bases to bases must be wrong - perhaps what was meant is that all invertible linear transformations map linearly independent sets to other linearly independent sets.

• @Display Name you mean linearly independent sets to other linearly independent sets. – Andrew Feb 14 at 22:18