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Commentary:
$\textbf{Exercise 15.}$ Suppose $v_1, \ldots, v_m$ is a linearly dependent list of vectors in $V$. Suppose also that $W \neq {0}$.
Prove that there exist $w_1, \ldots, w_m \in W$ such that no $T \in \mathcal{L}(V, W)$ satisfies $Tv_k = w_k$ for each $k = 1, \ldots, m$.
$\textbf{Solution 15.}$ If $v_1 = 0$, let $w_1$ be a nonzero vector in $W$. Then there does not exist a linear map $T \in \mathcal{L}(V, W)$ such that $Tv_1 = w_1$ (by 3.10).
If $v_1 \neq 0$, then by the linear dependence lemma (2.19) there exists $j \in {2, \ldots, m}$ and $c_1, \ldots, c_{j-1} \in \mathbb{F}$ such that $v_j = c_1v_1 + \cdots + c_{j-1}v_{j-1}.$ Let $w_k = 0$ for all $k = 1, \ldots, j-1$ and let $w_j$ equal any nonzero vector in $W$. Then the equation above and 3.10 imply that there does not exist a linear map $T \in \mathcal{L}(V, W)$ such that $Tv_k = w_k$ for all $k = 1, \ldots, j$.
$\textit{Commentary:}$ This exercise shows that if $v_1, \ldots, v_m$ is a linearly dependent list of vectors in $V$, and $W$ is a nonzero vector space, then there exist vectors $w_1, \ldots, w_m$ in $W$ such that no linear map from $V$ to $W$ sends each $v_k$ to the corresponding $w_k$.
The proof considers two cases. If $v_1 = 0$, then we can choose $w_1$ to be any nonzero vector in $W$. No linear map can send 0 to a nonzero vector, so there is no linear map $T$ with $Tv_1 = w_1$ in this case.
If $v_1 \neq 0$, then the linear dependence of $v_1, \ldots, v_m$ implies (by the linear dependence lemma) that some $v_j$ is a linear combination of the preceding vectors. We then choose $w_1, \ldots, w_{j-1}$ to be 0, and $w_j$ to be any nonzero vector in $W$. If a linear map $T$ satisfied $Tv_k = w_k$ for all $k = 1, \ldots, j$, then it would map the linear combination $c_1v_1 + \cdots + c_{j-1}v_{j-1}$ to 0 (because $w_1, \ldots, w_{j-1}$ are 0), but it would also map this same linear combination (which equals $v_j$) to the nonzero vector $w_j$. This is impossible, so no such linear map exists.
This exercise reinforces the understanding of the linear dependence lemma and its implications. It also provides practice in constructing examples where certain linear maps cannot exist.
$\textit{Examples:}$
1. Let $V = \mathbb{R}^3$ and $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$, $v_3 = (1, 1, 0)$. This is a linearly dependent list. Let $W = \mathbb{R}^2$ and $w_1 = (0, 0)$, $w_2 = (0, 0)$, $w_3 = (1, 0)$. Then no linear map $T : V \to W$ satisfies $Tv_k = w_k$ for $k = 1, 2, 3$.
2. Let $V = \mathcal{P}_2(\mathbb{R})$ and $v_1 = 1$, $v_2 = x$, $v_3 = x + 1$. This is a linearly dependent list. Let $W = \mathbb{R}$ and $w_1 = 0$, $w_2 = 0$, $w_3 = 1$. Then no linear map $T : V \to W$ satisfies $Tv_k = w_k$ for $k = 1, 2, 3$.
3. Let $V = \mathbb{C}^2$ and $v_1 = (1, 0)$, $v_2 = (i, 0)$, $v_3 = (1, i)$. This is a linearly dependent list. Let $W = \mathbb{C}$ and $w_1 = 0$, $w_2 = 0$, $w_3 = 1$. Then no linear map $T : V \to W$ satisfies $Tv_k = w_k$ for $k = 1, 2, 3$.
These examples demonstrate the principle in various vector spaces, including $\mathbb{R}^n$, polynomial spaces, and $\mathbb{C}^n$.
In each case, we have a linearly dependent list of vectors $v_1, \ldots, v_m$, and we construct a list of vectors $w_1, \ldots, w_m$ in a nonzero vector space $W$ such that no linear map from $V$ to $W$ maps each $v_k$ to the corresponding $w_k$.
The construction follows the proof: we set the $w_k