Question #337170

Given that the set S= ((1,0,0,0), (0,0,1.0), (5,1,11,0), (-4,0,-6,1)) is a basis of R*, and T= {(1,0,1,0), (0,2,0,3)} is linearly independent. Extend T to a basis of R.

1
Expert's answer
2022-05-05T06:30:08-0400

At first, we compute the determinant: A=(10000010511104061)A=\left(\begin{array}{llll}1&0&0&0\\0&0&1&0\\5&1&11&0\\-4&0&-6&1\end{array}\right). We receive: det(A)=1\det(A)=-1. Thus, the set SS indeed contains the basis (the vectors of SS are linearly independent and 55 vectors in R4\mathbb{R}^4 are always linearly dependent). Another basis, which can be received as an extension of TT, can be found in different ways. For example, we can use Gram-Schmidt orthogonalization. Another way is to consider the matrix: B=(10100203)B=\left(\begin{array}{llll}1&0&1&0\\0&2&0&3\end{array}\right). As we can see, the block (1002)\left(\begin{array}{llll}1&0\\0&2\end{array}\right) is non-degenerate. It means, that we can easily extend the matrix: B~=(1010020300100001)\tilde{B}=\left(\begin{array}{llll}1&0&1&0\\0&2&0&3\\0&0&1&0\\0&0&0&1\end{array}\right). The block (1001)\left(\begin{array}{ll}1&0\\0&1\end{array}\right) is also non-degenerate. det(B~)=2\det(\tilde{B})=2. Therefore, the respective vectors of matrix B~\tilde{B} are linearly independent. Thus, they are the basis.

Answer: (1,0,1,0)(1,0,1,0), (0,2,0,3)(0,2,0,3), (0,0,1,0)(0,0,1,0) and (0,0,0,1)(0,0,0,1) is the basis. It is received as the extension of TT.


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