Penney, Linear Algebra: Ideas and Applications (4th edition)

p.123 EXERCISES: 2.32, 2.33, 2.35, 2.36, 2.37,

2.39: you need to do two steps: (i) show that the vectors are linearly independent; (ii) show that any vector from \( \mathbb{R}^3 \) can be written as a linear combination of the basis vectors. Write down the linear combination explicitly.

2.46: in addition to what is asked in the textbook, write any linear combination of \( X \) and \( Y \) as a linear combination of \( A \) and \( B \) and vice versa.

Additional Problems.

1. Consider \( V = \mathrm{span}\{X_1, X_2, X_3\} \) where \( X_1, X_2, X_3 \) are linearly independent. (Hence \( A = \{X_1, X_2, X_3\} \) is a basis of \( V \) and \( \dim(V) = 3 \).)

   1. Let \( B = \{Y_1 = X_1 + 2X_2 - X_3,\ Y_2 = X_1 + X_2,\ Y_3 = 7X_2\} \). Show that \( B \) is also a basis for \( V \) by explicitly showing that: (i) \( B \) is linearly independent and, (ii) \( B \) spans \( V \). (For (ii), express any/every vector from \( V \) – not just \( X_1, X_2, X_3 \) – as a linear combination of the \( Y_i \)'s.
   2. Let \( C = \{Y_1 = X_1 + 2X_2 - X_3,\ Y_2 = X_1 - X_2,\ Y_3 = X_1 + X_2 + X_3,\ Y_4 = X_1 - 3X_2 + X_3\} \). Show that \( C \) is linearly dependent (and hence cannot be a basis for \( V \)).

   Do not use any theorem. Do the above from first principle, i.e. from scratch.

2. Given

   \[ \mathcal{L} = \left\{ X_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, X_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, X_3 = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \right\}, \quad \text{and} \quad Y = \begin{pmatrix} 1 \\ -3 \end{pmatrix}. \]

   Find constants \( c_1, c_2, c_3 \) such that \( Y = c_1 X_1 + c_2 X_2 + c_3 X_3 \) and \( c_1^2 + c_2^2 + c_3^3 \) is minimized.

   Hint: express \( c_1, c_2, c_3 \) in terms of (some) free variable(s). Then determine the free variable(s) so that \( c_1^2 + c_2^2 + c_3^3 \) is minimized. Recall your Calculus I.

   Remark/motivation about this problem. Clearly, the expression \( Y = c_1 X_1 + c_2 X_2 + c_3 X_3 \) is “inefficient” as we don’t need all the three vectors \( X_1, X_2, X_3 \) to express \( Y \) because \( X_3 \) can be written as a linear combination of \( X_1 \) and \( X_2 \) and hence can be thrown away. However, there might be “other notion(s)” of “efficiency”. For example, some applications might want to keep the coefficients as “small” as possible. In this problem, the “smallness” is measured by \( c_1^2 + c_2^2 + c_3^2 \).