# Linear Transformation and spanning set

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

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.

• 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. – layman Aug 30 '14 at 17:08
• 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. – Palu Aug 30 '14 at 17:34
• 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. – Palu Aug 30 '14 at 17:36
• 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. – Palu Aug 31 '14 at 14:13

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 $\dim(T(V))$ ?

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 $V$
• @Palu $\dim(T(V))\leq\dim(V)$ and yes – Belgi Aug 30 '14 at 16:57
• @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 $V$ – Belgi Aug 30 '14 at 17:14