Collection of Maths Problems

General independence

Task number: 2519

• Variant 1

  \(\{u,u+v,u+w\}\).

• Resolution

  We are seeking for coefficients \(a,b,c\in \mathbb R\) such that \(0=au+b(u+v)+c(u+w)=(a+b+c)u+bv+cw\).

  Now because \(u,v,w\) are linear independent it must hold that \(a+b+c=0\), \(b=0\) and \(c=0\) and thus also \(a=0\). Now we can see that the set \(\{u,u+v,u+w\}\) is linear independent.

• Variant 2

  \(\{u+v,u-v,w\}\).

• Result

  \(\{u+v,u-v,w\}\) is linear independent.

• Variant 3

  \(\{u+v,u-v,u+w,u-w\}\).

• Result

  \(\{u+v,u-v,u+w,u-w\}\) is linear dependent, for example by the use of coefficients \((1{,}1,-1,-1)^T\).

• Variant 4

  \(\{u+v,u+w,v+w\}\).

• Result

  \(\{u+v,u+w,v+w\}\) is linear independent.

• Variant 5

  \(\{u-v,u-w,v-w\}\).

• Result

  \(\{u-v,u-w,v-w\}\) is linear dependent, for example by the use of coefficients \((1,-1{,}1)^T\).

• Variant 6

  \(\{u-2v+w,3u+v-2w,7u+14v-13w\}\).

• Result

  \(\{u-2v+w,3u+v-2w,7u+14v-13w\}\) is linear dependent, for example by the use of coefficients \((5,-4{,}1)^T\).

• Variant 7

  \(\{u,2u,w\}\)

• Result

  \(\{u,2u,w\}\) is linear dependent, for example by the use of coefficients \((2,-1{,}0)^T\).

• Variant 8

  \(\{u,v+w\}\).

• Result

  \(\{u,v+w\}\) is linear independent.