Collection of Maths Problems

Independence in arithmetic vector spaces

Task number: 2520

• Variant 1

  \(X_1=\{(0{,}1,2{,}1)^T,(1{,}2,0{,}0)^T,(1{,}1,2{,}0)^T,(1{,}2,1{,}1)^T\}\).

• Resolution

  Similarly to the previous exercise we seek coefficients \(a,b,c\) and \(d\), such that the fllowing holds \(a(0{,}1,2{,}1)^T+b(1{,}2,0{,}0)^T+c(1{,}1,2{,}0)^T+d(1{,}2,1{,}1)^T=\mathbf 0\). This leas us to solving a homogeneous system of equations with a matrix: \[ \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 \\ 2 & 0 & 2 & 1 \\ 1 & 0 & 0 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 &-1 \\ 0 & 0 & 0 & -3 \\ \end{pmatrix} \]

  Now since the resulting matrix does not have free variables, the system has only trivial solution, and thus the set \(X_1\) is linear independent in \(\mathbb R^4\).

• Result

  \(X_1\) is linear independent is \(\mathbb R^4\).

  Similarly \(X_1\) is linear dependent in \(\mathbb Z_3^4\), now for example \((a,b,c,d)^T=(2{,}0,2{,}1)^T\), and finally in \(\mathbb Z_5^4\) is the set linear independent.

• Variant 2

  \(X_2=\{(1{,}1,2{,}0)^T,(1{,}2,1{,}1)^T,(0{,}1,2{,}1)^T,(1{,}1,0{,}0)^T\}\).

• Result

  \(X_2\) is linear dependent in \(\mathbb R^4\) by the use of coefficients \((1{,}2,-2,-3)^T\),

  it is linear dependent also in \(\mathbb Z_3^4\) as witnessed by \((1{,}2,1{,}0)^T\),

  and finally it is linear dependent in \(\mathbb Z_5^4\) as witnessed by \((1{,}2,3{,}2)^T\).

• Variant 3

  \(X_3=\{(1{,}0,2{,}0)^T,(2{,}1,0{,}2)^T,(0{,}2,2{,}1)^T,(2{,}2,1{,}1)^T\}\).

• Result

  \(X_3\) is linear independent in \(\mathbb R^4\),

  it is linear dependent in \(\mathbb Z_3^4\) as witnessed by \((0{,}1,2{,}2)^T\),

  and it is linear dependent in \(\mathbb Z_5^4\) as witnessed by \((2{,}0,1{,}4)^T\).