# Question 4

(ai)

We investigate:

\alpha(u + v) + \beta(u - v) + \gamma (u - 2v + w) = 0

That is:

(\alpha + \beta + \gamma)u + (\alpha - \beta - 2\gamma)v + \gamma w = 0

Since u,v,w are linearly independent we have \alpha + \beta + \gamma = 0, \alpha - \beta - 2\gamma = 0 and \gamma = 0. From the second equation we have \alpha = \beta, so 2\alpha = 0, giving \alpha = 0, so \beta = 0 too. So u + v, u - v and u - 2v + w are linearly independent.

(aii)

We investigate:

\begin{align*}\alpha(u + v - 3w) + \beta(u + 3v - w) + \gamma(v + w) &= (\alpha + \beta)u + (\alpha + 3\beta + \gamma)v + (-3\alpha - \beta + \gamma)w \\ &= 0\end{align*}

Again, we must have \alpha + \beta = 0, \alpha + 3 \beta + \gamma =0, -3\alpha - \beta + \gamma = 0. The first equation gives \beta= -\alpha, the second then gives \gamma = 2\alpha, then the third gives -3\alpha + \alpha + 2\alpha = 0. We therefore see that any triple (\alpha, \beta, \gamma) = (t,-t,2t) satisfies this system. Setting eg. t = 1 reveals that:

(u + v - 3w) - (u + 3v - w) + 2(v + w) = 0

so u + v - 3w, u + 3v - w, v + w are necessarily linearly dependent.

(bi)

Since \{v_1, v_2, \ldots, v_n\} are linearly independent we have:

\displaystyle \sum_{i = 1}^n \alpha_i v_i = 0

for some collection of n reals not all zero, \langle \alpha_i\rangle. We can write this as, since c_i \ne 0:

\displaystyle \sum_{i = 1}^n \frac {\alpha_i} {c_i} \left(c_i v_i\right) = 0

with \alpha_i not all zero. So the set \{c_1v_1, c_2v_2, \ldots, c_nv_n\} is linearly independent.

(bii)

We investigate:

\displaystyle \sum_{i = 1}^n c_i w_i

for some collection of n numbers \langle c_i\rangle not all zero.

We can expand this sum:

\displaystyle \sum_{i = 1}^n c_i w_i = \sum_{i = 1}^n c_i (v_i + v_1) = \left(\sum_{i = 1}^n c_i\right) v_1 + \sum_{i = 1}^n c_i v_i = \left(c_1 + \sum_{i = 1}^n c_i\right)v_1 + \sum_{i = 2}^n c_i v_i

Since \{v_1, v_2, \ldots, v_n\} are linearly independent we must have c_i = 0 for i \ge 2. We must also have:

\displaystyle 2c_1 + \sum_{i = 2}^n c_i = 0

we immediately find 2c_1 = 0, so c_1 = 0. So we necessarily have c_i = 0 for all i, and so \{w_1, w_2, \ldots, w_n\} is a linearly independent set.

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