Answer to Question #337170 in Linear Algebra for Darsheeta

Question

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.

Answer 1

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

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