existence of linear transformation doesn't hold

Question

Let V and W be two vector spaces over a field F. Does there exist a linear transformation that maps some vectors in a basis of V to $0$ and two vectors in the basis to a same nonzero vector in W? I think the answer is no. Let null space N(T) = {$v_1$, $v_2$, ..., $v_k$} and basis {$v_1$, $v_2$, ..., $v_k$, $v_{k+1}$, ... $v_n$}. Assume there exists $T(v_i) = T(v_j) \neq 0$ where $v_i, v_j \notin N(T)$. Then $T(v_i - v_j) = 0$. There exists a non-trivial linear combination $v_i - v_j = \sum_{m = 1}^{k}{a_m}{v_m}$. That is impossible. However，why existence of linear operator doesn't not hold here?

Answer 1

You can prescribe arbitrary values to vectors in a basis and extend the definition by linearity. So such linear transformations do exist.

Answer 2

Thanks @amd for answering the question. I finally understand what is wrong with my argument. In short, we can construct a vector basis from the basis of null space. But in some case, we cannot find a subset of the basis that is a basis of the null space. "The nullity of $T$ is greater than the number of vectors in the original basis that are mapped to zero". The new null space is larger than my original one, and some of the linear combinations of original pre-image become the basis of the new null space.

More specifically, let V and W be two vector spaces with dimension n and m over a field F. Let the basis of the null space $N(T)$ be {$v_1$, $v_2$, ..., $v_k$} and basis of $V$ be {$v_1$, $v_2$, ..., $v_k$, $v_{k+1}$, ... $v_n$}. For any linear transformation $T$, {$T(v_{k+1})$, ... $T(v_n)$} is a basis for range $R(T)$. When I construct a new transformation $S$ that maps $S(v_i) = S(v_j) \neq 0$ where $v_i, v_j \notin N(S)$, $Rank(S)$ is smaller than $Rank(T)$. By Rank–nullity theorem, new null space should have larger dimension than the original one. $v_i - v_j \notin \sum_{m = 1}^{k}{a_m}{v_m}$ doesn't hold for some nontrivial $a_m$ because $v_i - v_j$ now becomes a new basis vector.

Just imagine all basis vectors either map to zero or a nonzero vector.

Answer 3

Here is a minimal example of such a transformation: Let $V$ be three-dimensional. Fix a basis $\{\mathbf v_1,\mathbf v_2,\mathbf v_3\}$ of $V$ and choose a nonzero vector $\mathbf w\in W$. Define $T:V\to W$ as $$T:c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3\mapsto(c_1+c_2)\mathbf w.$$ This map is clearly linear and $T(\mathbf v_1)=T(\mathbf v_2)=\mathbf w$ and $T(\mathbf v_3)=0$ as required.

Where does your argument fail? To define $T$ in the first place, we must fix an ordered basis $\mathcal B = (\mathbf v_1,\dots,\mathbf v_n)$ of $V$. Without loss of generality, we can say that there’s some $m\ge2$ such that $T(\mathbf v_i)=0$ when $i\gt m$ and $T(\mathbf v_i)\ne0$ when $i\le m$. Let $S = \operatorname{span}\{\mathbf v_{m+1},\dots,\mathbf v_n\}$. Now, by hypothesis there is a pair of these basis vectors, say $\mathbf v_1$ and $\mathbf v_2$, such that $T(\mathbf v_1)=T(\mathbf v_2)\ne0$. But then $T(\mathbf v_1-\mathbf v_2)=0$, and so $\mathbf v_1-\mathbf v_2\in N(T)$. However, $\mathbf v_1-\mathbf v_2 \notin S$, so $S\subset N(T)$. This means that when you take a basis of $N(T)$ and extend it to a basis $\mathcal B'$ of $V$, as you do in your argument, this basis cannot be the same as $\mathcal B$: if $\mathbf v_1$ is an element of $\mathcal B'$, then $\mathbf v_2$ cannot be since it’s a linear combination of an element of $N(T)$ and $\mathbf v_1$, and similarly having $\mathbf v_2$ in $\mathcal B'$ excludes $\mathbf v_1$. Because of this, you can’t guarantee that there are two elements of $\mathcal B'$ with the same nonzero image under $T$, and the argument falls apart from there.

Going back to the example at top, the image of $T$ is obviously one-dimensional, so $\dim N(T)=2$. Extending any basis of $N(T)$ to a basis of $V$ involves adding only one more vector, so there’s no way to extend it in such a way that two of the basis vectors are mapped to the same nonzero vector by $T$.