Linear Algebra - Determining if a set is a basis for the indicated subspace

cheejudo

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Determine if{[1,1,1]T , [2,1,0]T} is a basis for the kernel of
1 -2 1
-1 2 -1
So, what I've tried is the following:Find the kernel of the given matrix. s*[2,1,0]T + t*[-1,0,1]T
So the spanning set for the kernel is {[2,1,0]^T, [-1,0,1]^T}
This spanning set is linearly independent, which will make it a basis for the kernel. But, I don't see how to determine if the original set is a basis.
 
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Linear Algebra - Determining if a set is a basis

[h=2][Linear Algebra] Determine if a set of vectors is a basis[/h]
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[FONT=&quot]Determine if
{[1,1,1]T , [2,1,0]T} is a basis for the kernel of
1 -2 1
-1 2 -1


So, what I've tried is the following:
Find the kernel of the given matrix. s*[2,1,0]T + t*[-1,0,1]T
So the spanning set for the kernel is {[2,1,0]^T, [-1,0,1]^T}
This spanning set is linearly independent, which will make it a basis for the kernel. But, I don't see how to determine if the original set is a basis.
[/FONT]
 
The kernel is the set of \(\displaystyle \begin{bmatrix}x \\ y \\ z \end{bmatrix}\) such that \(\displaystyle \begin{bmatrix}1 & -2 & 1 \\ -1 & 2 & -1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}x- 2y+ z \\ -x+ 2y- z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}\). We must have x- 2y+ z= 0 and -x+ 2y- z= 0. Of course, the second row of the matrix is just the negative of the first so those two equations are equivalent. We can write it as z= 2y- x so the vectors are or the form \(\displaystyle \begin{bmatrix}x \\ y \\ 2y- x\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}\).

The putative basis vectors for this space are \(\displaystyle \begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}\) and \(\displaystyle \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix}\).

It is sufficient to show that \(\displaystyle \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}= -1\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}+ \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix}\).
 
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