1
$\begingroup$

I have the following question below:

Let $V$ and $W$ be real vector spaces and $T:V\to W$ be a linear transformation such that $\ker(T) = {0} \subset V$. Let vectors $v_1,v_2,v_3,v_4$ belong to $V$, and $\{T(v_1), T(v_2)\}$ is a generating set for W. Is the set of vectors $v_1,v_2,v_3,v_4$ belonging to $V$ a generating set for $V$?

I found this question from an old book. The more current books I think use the word spanning set instead of generating set. I am not sure how to start this proof off. My initial guess is to assume

$w1$ = $T(v1)$, $w2$ = $T(v2)$, so $\{w1,w2\}$ is the spanning set in $W$. Such that some vector $wi$ in $W$ can be represented as $wi$ = a$w1$ + b$w2$. Then by linearity I guess that we can write $wi$ = $T( a$v1$+ b $v2$)$ .

From this point I am stuck to how to show that these 2 vectors are a spanning set for V, and if I can do that, I guess I can then add v3 and v4 to the first two and claim this is a spanning set for V, because a spanning set can have redundant vectors, this is not asking for a basis set, but a spanning set. Hope someone can help here.

Palu

$\endgroup$
1
$\begingroup$

First of all, mentioning the vectors $v_3$ and $v_4$ is pure smoke and mirrors, since we know nothing about them (they could even be zero). What is important is the fact that $\ker(T)=0$. This shows that $T$ is injective, so that the image $T(V)$ is isomorphic to $V$. Moreover the fact that $T(v_1)$ and $T(v_2)$ generate $W$ implies that $T$ is also surjective, so $T$ is an isomorphism, and $V$ and $W$ are isomorphic and thus $T(V)=W$. To show that $v_1$ and $v_2$ span $V$ take any vector $v \in V$, take it's image $T(v)$ in $W$ and decompose $T(v)=aw_1+bw_2$ and then take inverse images $v:=av1+bv_2$. This can only happen when $T$ is an isomorphism. An example where $T$ is not injective is given is T is the map $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by the matrix $ \left(\begin{array}{cc} 1&-2\\ 2&-4\\ \end{array}\right) $. The vectors $v_1=\left(\begin{array}{c} 1\\ 0\\ \end{array}\right)$ and The vectors $v_2=\left(\begin{array}{c} 0\\ 1\\ \end{array}\right)$ span the vectorspace $V=\mathbb{R}^2$ but their images by $T$ do not.

$\endgroup$
  • 1
    $\begingroup$ I just want to add a note. We don't necessarily know that $v_{1}$ and $v_{2}$ are linearly independent from each other, unless I'm mistaken. I don't think $\{ T(v_{1}), T(v_{2}) \}$ being a "generating set" necessarily means it is a basis. It doesn't affect your answer, but it is just an extra note. $\endgroup$ – layman Aug 30 '14 at 17:08
  • $\begingroup$ Thanks Nimda. So you have shown that only v1 and v2 span V and hence all of v1,v2,v3,v4 do not span V. Am I concluding correctly. $\endgroup$ – Palu Aug 30 '14 at 17:34
  • $\begingroup$ Hi user46944, yes you are correct, w1,w2 is a spanning set, not a basis and it is also correct that we don't know if these vectors, either vi or wi are independent. $\endgroup$ – Palu Aug 30 '14 at 17:36
  • $\begingroup$ Hi Nimba, yes since one takes inverse images, and get that a general vector v=av1+bv2, then we have a vector spanned by v1,v2 and then I guess one can by extension add in v3, v4, and hence have all of them become the spanning set (we can do this since we are looking for a spanning set and not a basis, we can allow these redundant vectors). So hence all 4 vectors we can say span the space V. Please let me your thoughts on this. $\endgroup$ – Palu Aug 31 '14 at 14:13
0
$\begingroup$

Hints: Since $W$ have a spanning set with two elements this means $\dim(W)\leq2$

Also note that $T$ is $1-1$ since it have a trivial kernel, this means $\dim(V)\leq\dim(W)\leq2$

Now - what in general can we say about $\dim(T(V))$ ?

Added in the comments:

Look at the set of inequalities we got: $$\dim(W)=\dim(T(V))\leq\dim(V)\leq\dim(W)$$ and thus $\dim(V)=\dim(W)$ and since $T$ is $1-1$ it is also onto. Now, since $\{T(v_{1}),T(v_{2})\}$ spans $W$ it have a subset that is a basis of $W$ (if $\dim(W)=2$ then it is simply $\{T(v_{1}),T(v_{2})\}$ and if $\dim(W)=1$ it is $\{T(v_{1})\}$ or $\{T(v_{2}\}$)

and $T^{-1}$ is also an isomorphism since $T$ is and it will map a basis to a basis so $\{v_{1},v_{2}\}$ or $\{v_{1}\}$ or $\{v_{2}\}$ is a basis for $V$

$\endgroup$
  • $\begingroup$ I am not sure about dim(T(V)), isnt that just dim(W). But from your second line, does that mean that in V that its dim is MAX 2? $\endgroup$ – Palu Aug 30 '14 at 16:55
  • $\begingroup$ @Palu $\dim(T(V))\leq\dim(V)$ and yes $\endgroup$ – Belgi Aug 30 '14 at 16:57
  • $\begingroup$ Hi and thanks. So I guess that dim(T(V)) is not in general all of dim(W), T(V)could be a subset of W. SO if dimension is say 2, then we can say that the number of vectors forming a basis in V is 2 vectors, but I can add the other 2 vectors, v3 & v4, and conclude that vectors, v1,v2,v3,v4 is a spanning set for V. $\endgroup$ – Palu Aug 30 '14 at 17:08
  • $\begingroup$ @Palu - In general no, but look at the set of inequalities we got: $\dim(W)=\dim(T(V))\leq\dim(V)\leq\dim(W)$ and thus $\dim(V)=\dim(W)$ and since $T$ is $1-1$ it is also onto. Now, since $\{T(v_{1}),T(v_{2})\}$ spans $W$ it have a subset that is a basis of $W$ (if $\dim(W)=2$ then it is simply $\{T(v_{1}),T(v_{2})\}$ and if $\dim(W)=1$it is $\{T(v_{1})\}$ or $\{T(v_{2}\}$) and $T^{-1}$ is also an isomorphism since $T$ is and it will map a basis to a basis so $\{v_{1},v_{2}\}$ or $\{v_{1}\}$ or $\{v_{2}\}$ is a basis for $V$ $\endgroup$ – Belgi Aug 30 '14 at 17:14
  • $\begingroup$ SO based on what you say, we have multiple possibilities, but these are all basis. The questions is asking for spanning set. SO if you say I cannot state v1,...,v4 is a spanning set, and I only look at the possibilities you state as basis, THEN I guess one is forced to conclude that v1,v2,v3,v4, all these together, is not a spanning set for V. $\endgroup$ – Palu Aug 30 '14 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.