Mullins Bolton
03/28/2023 · Elementary School

1. Consider the subset \( B: \equiv\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\} \subset \mathbb{C}^{3} \), where \[ \mathbf{b}_{1}: \equiv\left[\begin{array}{c}-1 \\ 0 \\ 1-\mathrm{i}\end{array}\right], \quad \mathbf{b}_{2}: \equiv\left[\begin{array}{c}2 \mathrm{i} \\ \mathrm{i} \\ -2 \mathrm{i}\end{array}\right], \quad \mathbf{b}_{3}: \equiv\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right] \] Let \( \mathbf{z}: \equiv\left[\begin{array}{l}z_{1} \\ z_{2} \\ z_{3}\end{array}\right] \in \mathbb{C}^{3} \) be an arbitrary vector. (a) Solve for the scalars \( c_{1}, c_{2}, c_{3} \in \mathbb{C} \). (b) Let \( l: B \rightarrow \mathbb{C}^{2} \) be the function defined by \( \quad l\left(\mathbf{b}_{1}\right): \equiv\left[\begin{array}{c}\mathrm{i} \\ 0\end{array}\right], \quad l\left(\mathbf{b}_{2}\right): \equiv\left[\begin{array}{c}0 \\ 1-2 \mathrm{~b}\end{array}\right], \quad l\left(\mathbf{b}_{3}\right): \equiv\left[\begin{array}{c}2 \\ -\mathrm{i}\end{array}\right] \) Explain why there exists a unique linear extension \( L: \mathbb{C}^{3} \rightarrow \mathbb{C}^{2} \) of \( l \)

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(a) The scalars \(c_1, c_2, c_3\) can be found by solving the system of equations derived from expressing \(\mathbf{z}\) as a linear combination of \(\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3\). (b) There exists a unique linear extension \(L: \mathbb{C}^3 \rightarrow \mathbb{C}^2\) of \(l\) because \(l\) is defined on a basis of a subspace and is linear.

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