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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) < $\infty$.

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  • $\begingroup$ 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. $\endgroup$
    – egreg
    Feb 14, 2019 at 22:34
  • $\begingroup$ @egreg Those are the assumption. $\endgroup$ Feb 14, 2019 at 22:46
  • $\begingroup$ 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. $\endgroup$
    – egreg
    Feb 14, 2019 at 22:52
  • $\begingroup$ @egreg I mean that "$V$ and $W$ having the same dimension" and "map $T: V \to W$ having {$0$} null space" are assumptions. $\endgroup$ Feb 14, 2019 at 22:57
  • $\begingroup$ 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. $\endgroup$ Feb 14, 2019 at 23:03

3 Answers 3

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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).

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  • $\begingroup$ I believe I just stated in that way: "null space is {$0$}"/trivial kernel and both spaces have the same dimensions. $\endgroup$ Feb 14, 2019 at 22:07
  • $\begingroup$ 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 $\endgroup$
    – Pedro
    Feb 14, 2019 at 22:12
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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.

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  • $\begingroup$ 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. $\endgroup$ Feb 14, 2019 at 22:10
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    $\begingroup$ @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. $\endgroup$
    – K.Power
    Feb 14, 2019 at 22:16
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    $\begingroup$ Nice! I just realized the importance of your first point. When setting a null space to be {$0$}, choice of operator T is limited. $\endgroup$ Feb 14, 2019 at 22:25
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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.

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  • $\begingroup$ @Display Name you mean linearly independent sets to other linearly independent sets. $\endgroup$
    – Andrew
    Feb 14, 2019 at 22:18

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