linear algebra

Problem Suppose $v_1, \dots, v_m$ is linearly independent in $V$ and $w \in V$. Show that $v_1, \dots, v_m, w$ is linearly independent if and only if $$w \thinspace \cancel{\in} \thinspace \textrm{span}(v_1, \dots, v_m)$$ Solution Let $a_0, a_1, \dots, a_m \in \mathbb{F}$ such that: $$0 = a_0w + a_1v_1 + a_2v_2 + \dots + a_mv_m$$ Then either $a_0 = 0$ which would imply $a_1 = a_2 = \dots = a_m = 0$ by the linear independence of $v_1, v_2, \dots, v_m$ or $a_0 \neq 0$ which would imply: $$w = \frac{a_1}{a_0}v_1 + \dots + \frac{a_m}{v_m}$$ i....

Problem Suppose $v_1, \dots, v_m$ is linearly independent in $V$ and $w \in V$. Prove that if $v_1 + w, \dots, v_m + w$ is linearly dependent, then $w \in \textrm{span}(v_1, \dots, v_m)$ Solution We prove the contrapositive - let $w$ $\cancel{\in}$ $\textrm{span}(v_1, \dots, v_m)$, then for $a_1,\dots,a_m \in \mathbb{F}$ we have: $$0 = a_1(v_1 + w) + a_2(v_2 + w) + \dots + a_m(v_m + w)$$ From which it follows that $-w\sum_{i=1}^{m}a_i = a_1v_1 + \dots + a_mv_m$ implying that $\sum_{i=1}^{m}a_i = 0$ for otherwise $w$ would be in the span of $v_1, \dots, v_m$: $$w = -\frac{a_1}{\sum_{i=1}^{m}a_i}v_1 - \dots - \frac{a_m}{\sum_{i=1}^{m}a_i}v_m$$ But $\sum_{i=1}^{m}a_i = 0$ implies that the left hand side in $-w\sum_{i=1}^{m}a_i = a_1v_1 + \dots + a_mv_m$ equals $0$, therefore by the linear independence of $v_1, \dots, v_m$ we must have $a_1 = a_2 = \dots = a_m = 0$....

Problem Prove or give counterexample: If $v_1, \dots, v_m$ and $w_1, \dots, w_m$ are linearly independent lists of vectors in $V$, then $v_1 + w_1, v_2 + w_2, \dots, v_m + w_m$ is linearly independent. Solution Let $w_1 = -v_1, w_2 = -v_2, \dots, w_m = -v_m$, then $v_1, v_2, \dots, v_m$ and $w_1, w_2, \dots, w_m$ are both linearly independent lists, but $v_1 + w_1, v_2 + w_2, \dots, v_m + w_m$ is just a list of zeros and is therefore clearly linearly dependent....

Problem Prove or give a counterexample: If $v_1, v_2, \dots, v_m$ is a linearly independent list of vectors in $V$ and $\lambda \in \mathbb{F}$ with $\lambda \neq 0$, then $\lambda v_1, \lambda v_2, \dots, \lambda v_m$ is linearly independent Solution By the linear independence of $v_1, v_2, \dots, v_m$ we must have that $$0 = a_1v_1 + a_2v_2 + \dots + a_mv_m = \frac{a_1}{\lambda}(\lambda v_1) + \frac{a_2}{\lambda}(\lambda v_2) + \dots + \frac{a_m}{\lambda}(\lambda v_m)$$ if and only if $a_1 = a_2 = \dots = a_m = 0$ by definition 2....

Problem Prove or give counterexample: If $v_1, v_2, \dots, v_m$ is a linearly independent list of vectors in $V$, then $$5v_1 - 4v_2, v_2, v_3, \dots, v_m$$ is linearly independent. Solution By the linear independence of $v_1, v_2, \dots, v_m$, we must have that $$0 = a_1v_1 + a_2v_2 + \dots + a_mv_m$$ $$=\frac{a_1}{5}\left(5v_1 - 4v_2\right) + \left(a_1\frac{9}{4} + a_2\right)v_2 + a_3 v_3 + \dots + a_m v_m$$ implies $a_1 = a_2 = \dots = a_m = 0$ which in turn implies $\frac{a_1}{5} = a_1\frac{9}{4} + a_2 = a_3 = \dots = a_m = 0$, thus the list $5v_1 - 4v_2, v_2, v_3, \dots, v_m$ is linearly independent if the list $v_1, v_2, v_3, \dots, v_m$ is....