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Find a linearly independent set of vectors that spans the same substance of $\mathbb{R}^3$ as that spanned by the vectors $\begin{bmatrix}2\\2\\-1\end{bmatrix}, \begin{bmatrix}-8\\-2\\5\end{bmatrix}, \begin{bmatrix}-3\\0\\2\end{bmatrix}$

So I put this matrix into RREF to get:

$\begin{bmatrix}1&0&1/2\\0&1&1/2\\0&0&0\end{bmatrix}$ , but how do I know if this is a linearly independent set of vectors? And also I tried to put in this for the answer and it said it was incorrect:

$\begin{bmatrix}1/2\\1\\0\end{bmatrix}, \begin{bmatrix}1/2\\0\\1\end{bmatrix}$

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    $\begingroup$ The set is linear dependent, because the third row of your RREF contains only zeros $\endgroup$ – imranfat Jul 11 '16 at 20:51
  • $\begingroup$ Ah, good point. Any idea why my answer is incorrect? $\endgroup$ – Yusha Jul 11 '16 at 20:52
  • $\begingroup$ Because these vectors aren't in the span of the given vectors. $\endgroup$ – imranfat Jul 11 '16 at 20:53
  • $\begingroup$ Interesting, so how can I find the vectors I need? $\endgroup$ – Yusha Jul 11 '16 at 20:54
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    $\begingroup$ True, so you can find two new lin independent vectors based on the first 2 $\endgroup$ – imranfat Jul 11 '16 at 21:01
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The vectors are linearly dependent, because that matrix has got a row of zeroes.

In general, if you find out that one of the vectors is a linear combination of some others, you can drop one of those and start againg from the vectors left.

In your example, having named $v_1,v_2,v_3$ those vectors, you have $v_1 + v_2 -3\,v_3\ =\ \bf{0}$, so you can drop one of the three, and the remaining couple are linearly independent if and only if they are not multiples.

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I personally feel that automatically putting things into "matrix form" and "row reducing" is too often a substitute for understanding what you are doing! A set of vectors, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$, is defined to be "linearly independent" if and only if the only solution to $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ is $a_1= a_2= \cdot\cdot\cdot= a_n= 0$.

The three vectors here are $(2, 2, -1)$, $(-8, -2, 5)$, $(-3, 0, 2)$ so the equation is $a(2, 2, -1)+ b(-8, -2, 5)+ c(-3, 0, 2)= (2a- 8b- 3c, 2a- 2b, -a+ 5b+ 2c)= (0, 0, 0)$ so we must have 2a- 8b- 3c= 0, 2a- 2b= 0, -a+ 5b+ 2c= 0. From 2a- 2b= 0, a= b. Putting that into the first equation, 2a- 8a- 3c= -6a- 3c= 0 so c= -2a. Putting b= a and c= -2a into the third equation -a+ 5a- 4a= 0a= 0. Then a(2, 2, -1)+ a(-8, -2, 5)- 2a(-3, 0, 2)= 0 so these vectors are linearly dependent. We can write a(2, 2, -1)= -a(-8, -2, 5)+ 2a(-3, 0, 2) and then divide by a: (2, 2, -1)= (-8, -2, 5)+ 2(-3, 0, 2). Since that vector can be written as a linear combination of the other two vectors, and those vectors [b]are[/b] independent (one is not a multiple of the other), the set {(-8, -2, 5), (-3, 0, 2)} is the largest subset of linearly independent vectors.

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  • $\begingroup$ a typo critically changes the nature of the result. the last vector is $(-3,0,2)$ which makes it linearly dependent set $\endgroup$ – gt6989b Jul 11 '16 at 21:10
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Hint:

the three given vectors are linearly dependent since: $$ \begin{bmatrix} -8\\-2\\5 \end{bmatrix}= -1 \begin{bmatrix} 2\\2\\-1 \end{bmatrix} +2\begin{bmatrix} -3\\0\\2 \end{bmatrix} $$ but any couple of these vectors are linearly independent, so any couples of them spans the same subpsace of $\mathbb{R}^3$.

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The RREF $\begin{bmatrix}1&0&1/2\\0&1&1/2\\0&0&0\end{bmatrix}$ tells you a lot about the the original set of column vectors.

The RREF has leading $1$'s in columns $1$ & $2$, so columns $1$ & $2$ in the original matrix, i.e. $\begin{bmatrix}2\\2\\-1\end{bmatrix}, \begin{bmatrix}-8\\-2\\5\end{bmatrix}$ form a set of linearly independent vectors with the same span as the original matrix.

Any linear relationship between columns in the RREF holds in the original matrix, e.g. it's easy to read "col $3 = \dfrac12$ col $1 + \dfrac12$ col $2$" in the RREF, which tells us $\begin{bmatrix}-3\\0\\2\end{bmatrix} = \dfrac12 \begin{bmatrix}2\\2\\-1\end{bmatrix} + \dfrac12 \begin{bmatrix}-8\\-2\\5\end{bmatrix}$

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Fact. Let $A$ be an $m\times n$ matrix. The nonzero rows of $\DeclareMathOperator{rref}{rref}\rref(A^\top)$ form a basis of the subspace of $\Bbb R^m$ spanned by the columns of $A$.

So, to solve your problem, put your vectors into the columns of a matrix $$ A= \left[\begin{array}{rrr} 2 & -8 & -3 \\ 2 & -2 & 0 \\ -1 & 5 & 2 \end{array}\right] $$ Row-reducing gives $$ \rref(A^\top)= \left[\begin{array}{rrr} 1 & 0 & -\frac{2}{3} \\ 0 & 1 & \frac{1}{6} \\ 0 & 0 & 0 \end{array}\right] $$ The fact above then implies that $\{\langle1,0,-2/3\rangle,\langle0,1,1/6\rangle\}$ is a basis of the subspace of $\Bbb R^3$ spanned by the columns of $A$.

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