115.C51 find a linearly independent set T so that T=S

[karush]

$\tiny{115.C51}$
find a linearly independent set T so that $\langle T\rangle =\langle S\rangle$
$S=\left\{
\left[\begin{array}{r}2\\-1\\2\end{array}\right],
\left[\begin{array}{r}3\\0\\1\end{array}\right],
\left[\begin{array}{r}1\\1\\-1\end{array}\right],
\left[\begin{array}{r}5\\-1\\3\end{array}\right]
\right\}$
make matrix A and derive RREF(A) to find pivot columns
$A=\left[
\begin{array}{rrrr}
2 & 3 & 1 & 5 \\
-1 & 0 & 1 & -1 \\
2 & 1 & -1 & 3
\end{array} \right]
\quad
\text{RREF}(A)=\left[
\begin{array}{rrrr}
1 & 0 & -1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0
\end{array} \right]$
\The pivot columns are observed at $C_1$ and $C_2$ thus we have
$T=\left\{
\left[\begin{array}{r}2\\-1\\2\end{array}\right],
\left[\begin{array}{r}3\\0\\1\end{array}\right]
\right\}$
then $\langle T\rangle =\langle S\rangle$
ok I think this is correct but I just followed a similar example
not sure just why this would be T=S when it looks like a subset
also, not up on the all the standard notatons for these type of problems
anyway mahalo much for any help
https://dl.orangedox.com/5LxJq55fJyIgYJE0y0

 

[jvicens]

Hello,

Thank you for sharing your work on this problem. Your solution is correct. To find a linearly independent set T that generates the same subspace as S, we can use the row reduction method to find the pivot columns of the given matrix A. These pivot columns will form the basis for the subspace generated by S. In this case, we have pivot columns at $C_1$ and $C_2$, which correspond to the first two columns of A. Therefore, the vectors in S corresponding to these pivot columns, namely $\left[\begin{array}{r}2\\-1\\2\end{array}\right]$ and $\left[\begin{array}{r}3\\0\\1\end{array}\right]$, will form a linearly independent set T that generates the same subspace as S.

As for your question about why T=S, it is because T is a subset of S. In other words, the vectors in T are also present in S, but T may not contain all the vectors in S. In this case, T only contains two out of four vectors in S. However, these two vectors are sufficient to generate the same subspace as S.

Also, the standard notation for this type of problem is to use the angle brackets $\langle \rangle$ to denote the subspace generated by a set of vectors. So $\langle T\rangle$ represents the subspace generated by the vectors in T.

I hope this helps clarify any confusion. Keep up the good work!