# Inner product of linearly independent rows of $m \times n$ matrix with $n$ linearly independent vectors gives $m$ independent vectors

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.

• 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 Dec 6, 2019 at 16:01

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.
• Those tuples should have $n$ components. Shouldn't it be $(a_{j1},...,a_{jn})$ if I didn't understand wrongly?
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.