I have a Field K and a k-vector space Homorphismus, $ \phi : {V} \times {W}$ and ${n \in N_0}$ and a n-Tupel $({s_1} ,\dots, {s_n})$ in ${V}$ (a) $(\phi(s_1,\dots ,\phi(s_n))$ $\textbf{linearly independent}$ in $ {V}$ $\Rightarrow$ $({s_1}, \dots ,{s_n})$ $\textbf{linearly independent}$ in ${V}$

(b) I have a Filed K and a Set X ,${n,m \in N_0}$,and a n-Tupel $({f_1} ,\dots, {f_n})$ in ${Map(X,K)}$, and a m-Tupel $({x_1} ,\dots, {x_m})$ in ${X}$ then :

$((f_1(x_1),\dots,f_1(x_m)),\dots,(f_n(x_1),\dots,f_n(x_m)))$ $\textbf{linearly independent}$ $\Rightarrow$ $({f_1} ,\dots, {f_n})$ $\textbf{linearly independent}$ in ${Map(X,K)}$,

to answer (b) I did define a vector space Homorphismus structure and used (a) let :

$\phi: Map(X,K)\rightarrow {K^m} , f_i \mapsto (f_i(x_1),\dots,f_i(x_m)) $ and I did prove that my structure can be vectorspace Homorphismus and then used to (a) to prove (b) ( is that correct ? )

(c) let $ k \in Z$, $f_k: R \rightarrow R ,x\mapsto \sin(kx)$ and $g_k: R \rightarrow R ,x\mapsto \cos(kx)$

the Question is if $(f_1,f_2,g_1,g_2)$ $\textbf{linearly independent}$ in Map(R,R) ?

$\textbf{my idea}$

to prove (c) i will use (b) , let m=3 and $x_1=\pi$ and =$x_2= 2x_1$, and $x_3=3x_1$

then i must prove that $((\sin(\pi),\sin(2\pi),\sin(3\pi),(\sin(2\pi),\sin(4\pi),\sin(6\pi)(\cos(\pi),\cos(2\pi),\cos(3\pi)),(\cos(2\pi),\cos(4\pi),\cos(6\pi)))$ =((0,0,0),(0,0,0,0),(-1,1,-1),(1,1,1)) and it is $\textbf{linearly independent}$ then according to (b) is $(f_1,f_2,g_1,g_2)$ $\textbf{linearly independent}$ in Map(R,R).

Is my idea correct ? in (b) and (c)


1 Answer 1


Your ideas are basically correct.

For (c), your $4$ vectors only span $2$ dimensions (include the zero vector twice(!)), so it's not linearly independent.

We need another choice of the $x_i$'s. Also, $m=3$ seems insufficient for proving the independency of $4$ vectors.

What about taking e.g. $x_1=0,\ x_2=\pi/2, x_3=\pi, x_4=3\pi/2$ or something along these lines? It's important to arrive at a linearly independent family of vectors in $\Bbb R^4$.

  • $\begingroup$ yeah that would be correct too , but the reason i did choose , m=3, is , according b the choice of x is not restricted , and the the proof must be coorect for all m and x, but can you explain,why would i choose m=4 and not 3 ? $\endgroup$
    – Mohbenay
    May 22, 2017 at 6:03
  • $\begingroup$ I found out that they are linearly dependent , how can i say then that (f1,f2,g1,g2) are linearly dependent , i have only $A \Rightarrow B$ and not $A \Leftrightarrow B$ $\endgroup$
    – Mohbenay
    May 22, 2017 at 8:43
  • $\begingroup$ They are independent. Make another choice. To derive that $4$ vectors are independent, we need their projection to a space of dimension at least $4$. $\endgroup$
    – Berci
    May 27, 2017 at 20:09

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