Find an example for each of the following, or explain why no example can be found:
a) a linear map $\phi_3 : M_{2\times 2}(\mathbf{R}) \to \mathbf{R}_4$ and matrices $\ell_1, \ldots, \ell_4 \in M_{2\times2}(\mathbf{R})$ that are linearly dependent such that $\phi_3(\ell_1), \ldots, \phi_3(\ell_4)$ are linearly independent.
I believe an example cannot be found for this statement, but what would a proof for this be?
b) a linear map $\phi_4 : M_{2\times2}(\mathbf{R}) \to \mathbf{R}^4$ and matrices $k_1, \ldots, k_4 \in M_{2\times2}(\mathbf{R})$ that are linearly independent such that $\phi_4(k_1), \ldots, \phi_4(k_4)$ are linearly dependent.
I believe an example for this can be found, my example is \begin{align*} m_1 &= \left( \begin{smallmatrix} {1} & {1}\\ {0} & {0} \\\end{smallmatrix} \right) \\ m_2 &= \left( \begin{smallmatrix} {0} & {1}\\ {0} & {0} \\\end{smallmatrix} \right) \\ m_3 &= \left( \begin{smallmatrix} {0} & {0}\\ {1} & {0} \\\end{smallmatrix} \right) \\ m_4 &= \left( \begin{smallmatrix} {0} & {0}\\ {0} & {1} \\\end{smallmatrix} \right) \end{align*} This set of matrices is clearly linearly independent, and
$$\phi_4: \left( \begin{smallmatrix} {a} & {b}\\ {c} & {d} \\\end{smallmatrix} \right) \mapsto (a-b,c+d, 2c, 2d)$$
gives $\phi_4(m_1) = (0,0,0,0)$. So as we already have a 0 vector, it does not matter what $\phi_4(m_2),\phi_4(m_3), \phi_4(m_4)$ are, as we have already attained a 0 vector so any integer multiplied by ϕ(m1) satisfies the condition for linear dependence.
Is this example sufficient?
$l_1,l_2,l_3, l_4$ are linearly dependent
there exists some not trivial set of scalars $(c_1,c_2,c_3,c_4)$ such that $c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4 = 0$
$\phi(l)$ is a linear map
$\phi(c_1l_1 + c_2l_2 +c_3l_3 + c_4l_4) = c_1\phi(l_1) + c_2\phi(l_2) + c_3\phi(l_3) + c_4\phi(l_4) = 0$
$\phi(l_1),\phi(l_2),\phi(l_3),\phi(l_4)$ are linearly dependent.
2) Your example seems fine, but you could probably find something even simpler.