Collection of Maths Problems

Inclusion of vector spaces

Task number: 2542

• Variant 1

  \(U_1=\mathcal L((4{,}1,0{,}3,4{,}0,0)^T,(4{,}3,1{,}0,2{,}3,1)^T,(4{,}1,4{,}0,3{,}2,4)^T,\\\strut\qquad (2{,}4,1{,}4,4{,}3,1)^T,(0{,}4,3{,}2,2{,}4,3)^T)\)

  \(V_1=\{(x_1,…,x_7)^T\in\mathbb Z_5^7: x_1+3x_2+ x_3+2x_4+3x_5+ x_6+2x_7=0,\\\strut\qquad 3x_1+4x_2+3x_3+ x_4+4x_5+2x_6+4x_7=0,\ 2x_1+ x_2+4x_3 +4x_5 +2x_7=0\}\)

• Resolution

  The dimension of a subspace is smaller than the dimension of its superspace. Since we have calculated the dimensions earlier in one of the previous problems, we may exclude the case \(V_1 \subset U_1\). It remains to verify whether the space are in the opposite inclusion or whether they are incomparable. It suffices to verify whether \(\dim(\mathcal L(U_1\cup V_1))=\dim(V_1)=4\).

  Alternatively we may try to express the vectors of the basis of the smaller space as a linear combination of the vectors from the other basis (this is in principle the same).

  \def\dall{\multicolumn{1}{c}{ }} \( \begin{array}{|rrrr|rrrrrrr|} \cline{5-11} \dall& & & & 2 & 1 & 0 & 0 & 0 & 0 & 0\\ \dall& & & & 1 & 0 & 2 & 1 & 0 & 0 & 0\\ \dall& & & & 1 & 0 & 1 & 0 & 1 & 0 & 0\\ \dall& & & & 1 & 0 & 4 & 0 & 0 & 3 & 1\\ \noalign{\hrule} 2 & 2 & 2 & 3 & 1 & 2 & 3 & 2 & 2 & 4 & 3\\ 1 & 0 & 2 & 1 & 0 & 1 & 1 & 0 & 2 & 3 & 1\\ 0 & 3 & 1 & 1 & 0 & 0 & 1 & 3 & 1 & 3 & 1\\ \noalign{\hrule} \end{array} \)

  The rows of the first matrix are the coordinates of the vectores of the basis of \(U_1\) with respect to the basis \(V_1\) (both bases have been calculated earlier). Observe thet these coordinates na be straightforwardly determined from the 2., 4., 5. and 7. component of the vector, argue why.

  To extend the basis we start with any basis of the smaller space and insert vectors of the larger until we get dimension 4.

• Result

  It holds that \(U_1 \subset V_1\).

  This inclusion can be seen in an easier way: all vectors that generate \(U_1\) satisfy equations of the definition of \(V_1\).

• Variant 2

  \(U_2=\mathcal L((1{,}2,4{,}2,3{,}1,2)^T,(2{,}3,4{,}1,2{,}1,3)^T,(3{,}4,1{,}1,4{,}1,4)^T,\\\strut\qquad (4{,}0,2{,}3,3{,}4,1)^T,(4{,}3,1{,}3,2{,}3,2)^T)\)

  \(V_2=\{(x_1,…,x_7)^T\in\mathbb Z_5^7: x_1+2x_2+ x_3 + x_5+2x_6+3x_7=0,\\\strut\qquad 4x_1+2x_2+ x_3+3x_4+2x_5 + x_7=0,\ x_1+ x_2+3x_3 + x_6 =0\}\)

• Result

  \(V_2 \subset U_2\).