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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.

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    $\begingroup$ Try a proof by contradiction. If they are linearly dependent, we have a non trivial solution. Then collect terms in each of the v_i $\endgroup$ – Calvin Lin Dec 6 '19 at 16:01
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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.

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  • $\begingroup$ Those tuples should have $n$ components. Shouldn't it be $(a_{j1},...,a_{jn})$ if I didn't understand wrongly? $\endgroup$ – Shatabdi Sinha Dec 6 '19 at 16:54
  • $\begingroup$ Yes, thanks for noticing. $\endgroup$ – Martin Argerami Dec 6 '19 at 17:19
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Let $$\sum_{i=1}^m\lambda_iw_i=0$$ with $\lambda_i\in K $. You want to prove that all $\lambda_i$ must be $0$.

Expand:$$\sum_{i=1}^m\lambda_i\sum_{j=1}^na_{ij}v_j=0$$

Rewrite:$$\sum_{j=1}^n\left(\sum_{i=1}^m\lambda_ia_{ij}\right)v_j=0$$ Now use the fact that the $v_j$ are independent and then the fact that the $(a_{ij})$ are independent.

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