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Look at the following exercise taken from a Linear Algebra I exam:

Let $X:= \{v_1,v_2,v_3,v_4\} \subseteq \mathbb R^4$ and $Y:= \{w_1,w_2,w_3\} \subseteq \mathbb R^3$ be $\mathbb R$-bases. Look at the subspaces $V = \langle v_1, v_2, v_3 \rangle$ and $W = \langle w_1, w_2 \rangle$. Let $f_M: \mathbb R^4 \to \mathbb R^3 $ be a $\mathbb R$-homomorphism with $f_M(V) \subseteq W$ which is, with respect to the bases $X, Y$, given by $$M= \begin{pmatrix} m_{11} &m_{12} &m_{13}&m_{14} \\ m_{21} &m_{22} &m_{23}&m_{24} \\ m_{31} &m_{32} &m_{33}&m_{34} \end{pmatrix}$$ a) Show that there is a unique $\mathbb R$-homomorphism $\widetilde{f_M} : V \to W$ with $\widetilde{f_M}(v) = f_M(v)$ for all $v\in V$.
b) Show that the transformation matrix of $\widetilde{f_M}$ with respect to the bases $\{v_1, v_2, v_3\}$ and $\{w_1, w_2\}$ is given by $$M= \begin{pmatrix} m_{11}&m_{12}&m_{13} \\ m_{21}&m_{22}&m_{23} \end{pmatrix}$$

Now for a), I don't know if this is trivial or not but the existence of $\widetilde{f_M}$ is trivial since it is defined for all $v\in V$ by $f_M$ which is a $\mathbb R$-homomorphism. Uniqueness is trivial as well, or am I missing something here?

Looking at b):
It holds that $f_M(v_1) = m_{11} w_1 + m_{21} w_2 + m_{31} w_3$. But since $f_M(V) \subseteq W$ it follows that $m_{31} = 0$ (analogous for $v_2$ and $v_3$) and we're done. I'm not sure about this step, if it is legit to conclude $m_{31} = 0$.

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In your answer for a) you let your intuition work, and in this case your intuition was right, but in a case where the bases of $V$ and $W$ would have been different your intuition would have failed you. The setup was deliberatily chosen in order to make the exercise easy but still a rigorous solution was expected. As for b) the uniqueness is not proven at all. We have the following setup

We have the following commutative diagram of lineair transformations: $$ \begin{array}{} V & \stackrel{\widetilde{f_M}}{\longrightarrow} & W\\ \downarrow i_V & & \uparrow p_W\\ \Bbb R^4 & \stackrel{f_M}{\longrightarrow} & \Bbb R^3 \end{array} $$

The diagram being commutative means that when you go from $V$ to $W$ in two ways you have to obtain the same result every which way go go. The big unknown is the projection $p_W$ from a three dimensional space into a $2$ dimensional one. Your choice of mapping $w_1 \mapsto w_1$ and $w_2 \mapsto w_2$ is fine, but nothing proves that this is the only way. That is what we will do now. Let us first represent the two paths as matrices :

  1. Path 1 : Directly from $V$ to $W$ we will represent this by a matrix $\begin{pmatrix}m_{1, 1} & m_{1, 2} & m_{1, 3}\\m_{2, 1} & m_{2, 2} & m_{1, 3}\end{pmatrix}$ representing the map $\widetilde{f_M}$.

  2. Path 2: We start with the injection $i_V$ with matrix $\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}$, followed by the given linear transformation $f_m$ with matrix $\begin{pmatrix}m_{1, 1} & m_{1, 2} & m_{1, 3} & m_{1, 4}\\m_{2, 1} & m_{2, 2} & m_{2, 3} & m_{2, 4}\\m_{3, 1} & m_{3, 2} & m_{3, 3} & m_{3, 4}\end{pmatrix}$ and next the unknown projection $p_W$ with unknown matrix elements $\begin{pmatrix}a & b & c\\d & e & f\end{pmatrix}$ so the compositioin of $p_W \circ f_M \circ i_V$ (note the writing from right to left): $$ \begin{pmatrix}a & b & c\\d & e & f\end{pmatrix} \begin{pmatrix}m_{1, 1} & m_{1, 2} & m_{1, 3} & m_{1, 4}\\m_{2, 1} & m_{2, 2} & m_{2, 3} & m_{2, 4}\\m_{3, 1} & m_{3, 2} & m_{3, 3} & m_{3, 4}\end{pmatrix} \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix} $$ Resulting in $$ \begin{pmatrix}a m_{1, 1} + b m_{2, 1} + c m_{3, 1} & a m_{1, 2} + b m_{2, 2} + c m_{3, 2} & a m_{1, 3} + b m_{2, 3} + c m_{3, 3}\\d m_{1, 1} + e m_{2, 1} + f m_{3, 1} & d m_{1, 2} + e m_{2, 2} + f m_{3, 2} & d m_{1, 3} + e m_{2, 3} + f m_{3, 3}\end{pmatrix}$$ So the diagram is commutative if and only if $a=1, b=0, c=0, d=0, e=1, f = 0$. Your "intuitive" solution, but now proven rigorously.

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