This question is a supplementary problem of Schaum's outlines Linear Algebra.
The question says:
Suppose $(a_{11},..,a_{1n}), \dots \dots , (a_{m1},..,a_{mn}) $ are
linearly independent vectors in $K^n$, and suppose $v_1, v_2, \dots,
v_n$ are linearly independent vectors in a vector space $V$ over $K$.
Show that the vectors$w_1 = a_{11}v_1 + \dots +a_{1n}v_n, \dots , w_m = a_{m1}v_1 + \dots +a_{mn}v_n$
are also linearly independent.
Being a beginner, I am unable to understand how to start the proof. I tried using the definition of independence of vectors and i know a bit about rowspace but I really don't know what works here.
Best Answer
To test if $w_1,\ldots,w_m$ are linearly independent, we set a linear combination equal to zero, and set to show that the coefficients have to be zero. So we assume $$ 0=\sum_{j=1}^mb_jw_j. $$ Then $$ 0=\sum_{j=1}^mb_jw_j=\sum_{j=1}^mb_j\sum_{k=1}^na_{jk}v_k=\sum_{j=1}^m\sum_{k=1}^nb_ja_{jk}v_k=\sum_{k=1}^n\left(\sum_{j=1}^mb_ja_{jk}\right)v_k. $$ Because of the linear independence of $v_1,\ldots,v_n$, we get that $$ \sum_{j=1}^mb_ja_{jk}=0,\ \ \ k=1,\ldots,n. $$ These are precisely the $n$ entries of $$ \sum_{j=1}^mb_j(a_{j1},\ldots,a_{jn}), $$ so $$ \sum_{j=1}^mb_j(a_{j1},\ldots,a_{jn})=0. $$ Now the linear independence of the vectors $(a_{j1},\ldots,a_{jn})$ gives you that $b_1=\cdots=b_m=0$. So $w_1,\ldots,w_m$ are linearly independent.