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my question is how to determine whether a vector $w$ is in the $span\{v_1, v_2, v_3\}$. In this case:

$w = \begin{bmatrix} 9 \\ 6 \\ 1 \\ 9 \\ \end{bmatrix} $ and $v_1 = \begin{bmatrix} 1\\2\\-1\\1 \end{bmatrix}$ , $v_2 = \begin{bmatrix} 2\\-1\\1\\0 \end{bmatrix}$ , $v_3 = \begin{bmatrix} 1\\2\\0\\3 \end{bmatrix}$

My understanding so far is that I must see if $w$ can be written as a linear combination of $v_1, v_2,$ and $v_3$. To do this, I row reduced the augmented matrix of the 4 vectors to the identity matrix $I_4$

$$I_4 = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{array}\right]$$

but from here I am confused. I'm just trying to better understand $span$, both in a problem like this and in a conceptual sense.

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  • $\begingroup$ Take a close at that last line....that should raise some eyebrows.... $\endgroup$
    – imranfat
    Apr 28 '16 at 20:41
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Hint: Let $w = \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3$. Then you will have four equations in three unknown quantities. Solve them.

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If $\begin{bmatrix}v_1,v_2,v_3|w\end{bmatrix}$ can be row reduced to $I_4$, then {$v_1,v_2,v_3,w$} form a basis for $\mathbb R^4$.

What are the implications of that? The 4 vectors are linearly independent, $w$ cannot be formed as a combination of $v_1,v_2,v_3$ and $w$ is not in the space spanned by $v_1,v_2,v_3$.

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You have all the four vectors $(w, v_1, v_2, v_3)$. To decide if $w \in Spam\{v_1, v_2, v_3\}$, you need to check if w is written as a linear combination of $v_1, v_2, v_3$, ie, you need to check if there exist $a_1, a_2, a_3 \in \mathbb{R}$ such that $w = a_1v_1 + a_2v_2 + a_3v_3$.

Then, you can solve a system with 4 equations. If you find a solution, then $w \in Spam\{v_1, v_2, v_3\}$,

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Hint 1: If you have $v_1, \dots, v_n \in \mathbb{R}^n$ they are linear independent, if and only if $\det (v_1, \dots, v_n) \neq 0$.

Hint 2: If $\lbrace v_1, v_2, v_3, v_4\rbrace$ is linear independent, then $w \in \text{span} \lbrace v_1, v_2, v_3\rbrace$ if and only if set $\lbrace v_1, v_2, v_3, w\rbrace$ is linear dependent.

Solution: Check, if $v_1, v_2, v_3$ are linear independent. To do that, you can check, if $\dim \text{span}\lbrace v_1, v_2, v_3 \rbrace = 3$, but dimension is size of biggest minor with non zero determinant.

Cause (take 3 first value from $v_i$ vectors)

$$ \det \left(\begin{split} &1 & 2 & (-1) \\ &2 & (-1)~~~ & 2 \\ &(-1)~~~ & 1 & 0 \end{split}\right) \neq 0 $$

$v_1, v_2, v_3$ are linear independent. If they aren't take base $A$ of $\text{span} \lbrace v_1, v_2, v_3 \rbrace$.

Check, if $A \cup \lbrace w \rbrace$ is linear independent, same way. In this way

$$ \det (v_1, v_2, v_3, w ) = det \left( \begin{split} &9 & 1 & 2 & 1\\ &6~~ & 2 & (-1)~~ & 2\\ &1 & (-1)~~ & 1 & 0\\ &9 & 1 & 0 & 3\\ \end{split} \right) = d $$

$$w \in \text{span} \lbrace v_1, v_2, v_3 \rbrace \Longleftrightarrow d = 0$$

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  • $\begingroup$ Can you elaborate on your hints more? For hint one - is this true for all matrices or just a square matrix? For hint two - are $\{v_1, v_2, v_3\}$ are linearly independent by themselves? $\endgroup$
    – brdeav39
    Apr 29 '16 at 3:14
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What you proved is that $\vec{w} \in \text{span}\{v_1, v_2, v_3\}$ if and only if the last column of $I_4$ is in the span of its first 3 columns. Is it?

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You can analyze a matrix as a linear transformation(function) which maps from vector space to another vector space or to the same one . If you want to check whether any vector will be spanned by given set of vectors , you have to look at the image of that transformation (function) . If the given vector w lies in the image of that function , then the given set of vectors successfully spans the vector w . Well in this case , if your row reduced echelon form is calculated without error , then the set will not span the vector since the last row reveals that 0.v1 + 0.v2 + 0.v3 = 1 which is not possible . Hence the given vectors doesn't span vector w . Hope this helps .

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