Question #229867
Let { (1,1,1,1) , (1,2,1,2) } be a linearly independent sbuset of a vector space V4. Extend it to a basic of V4
1
Expert's answer
2021-08-26T18:22:57-0400

It is enough to find a matrix L=(11111212x1x2x3x4y1y2y3y4)L=\left(\begin{array}{llll}1&1&1&1\\1&2&1&2\\x_1&x_2&x_3&x_4\\y_1&y_2&y_3&y_4\end{array}\right) wit nonzero determinant. Set x1=0x_1=0, x2=0x_2=0, x3=1x_3=1, x4=0x_4=0; y1=0y_1=0, y2=0y_2=0, y3=0y_3=0 , y4=1y_4=1. The determinant of the matrix is: det(L)=det(A)det(B)det(L)=\det(A)\det(B), where A=(1112)A=\left(\begin{array}{ll}1&1\\1&2\end{array}\right) and B=(1001)B=\left(\begin{array}{ll}1&0\\0&1\end{array}\right). det(A)=1det(A)=1, det(B)=1det(B)=1. det(L)=1det(L)=1. Thus, the basis is: (1111)(1\,\,1\,\,1\,\,1), (1212)(1\,\,2\,\,1\,\,2), (0010)(0\,\,0\,\,1\,\,0), (0001)(0\,\,0\,\,0\,\,1).


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