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Give algebraic and geometric descriptions of $\operatorname{Span} \{ a_1, a_2, a_3, a_4 \}$ where

$a_1 = (1, -1, -2), a_2 = (3, -3, -1), a_3 = (2, -2, -4), a_4 = (2, -2, 1)$

So far, I have: $$ \begin{matrix} \;\;\,1 & \;\;\,3 & \;\;\,2 & \;\;\,2 \\ -1 &-3 &-2 &-2\\ -2 & -1 &-4 & \;\;\,1 \\ \end{matrix} $$ Though, I feel like I'm missing a column. What should this system be equal to?

3 Answers 3

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$\begin{pmatrix} 1 & 3 & 2 & 2\\ -1& -3& -2& -2\\ -2& -2& 4& 1 \end{pmatrix}$

only use elementary row operation,we can get

$ \begin{pmatrix} 1 &0 &2 &-1 \\ 0&1 & 0 &1 \\ 0&0 &0 &0 \end{pmatrix}$

then,$a_1=2a_3$,and $a_4=-a_1+a_2$

$\operatorname{Span} \{ a_1, a_2, a_3, a_4 \}=\operatorname{Span} \{ a_1, a_2\}$

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    Great! So this would be a sufficient algebraic solution. The geometric solution would be the plane that contains vectors a1, a2, and 0; more specifically, the plane that contains the line in R3 through vectors a1 and 0 and the line in R3 through vectors a2 and 0. I was confused as to the easiest way to find the algebraic solution. Thank you so much.2012-09-10
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    what confused you?2012-09-10
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    I missed the obvious--I didn't realize adding and subtracting the results of Gaussian elimination would help me prove linear dependence.2012-09-10
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    if you think my answer is right,you can select it as the best answer of your question.2012-09-10
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To start, notice that $a_1=2a_3$, thus we may eliminate $a_3$ as redundant. Can you see a relation between the remaining vectors?

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    They each increment by 1 or -1, depending on the sign...2012-09-10
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    Remember, if we can write one vector as a linear combination of the other two, then that vector adds nothing to the span...there is a linear relationship between $a_1=(1,-1,-2), a_2=(3,-1,-1),$ and $ a_4=(2,-2,1)$. What is it?2012-09-10
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    I'm sorry, but I only see a relationship between a1 and a3. The others don't seem to have a simple constant to equal one another.2012-09-10
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    I know that if a1 = 2a3, then the two vectors are *not* linearly independent, correct?2012-09-10
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    That is correct. And if $a_1=ca_2+da_4$ for some constants $c,d$, then $a_1$ is dependent on $a_2$ and $a_4$. What are the constants in this case?2012-09-10
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    Ah, my mistake. I didn't understand the term linear combination fully. If ALL are linearly dependent... then geometrically this would just be a line in R2 space.2012-09-10
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    That would be the case if they were ALL constant multiples of each other, but that is not the case here. There are two linearly independent vectors; find them and they will determine a plane.2012-09-10
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    Ok. Is there a fast way to compute linear independence? For example, setting up a matrix and then realizing there is no solution when 0 = a number?2012-09-10
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    I calculated the matrix of a2, a4 = a1 and got 1 0 | 3, 0 1 | -1, 0 0 | 0, but I'm not sure what this means exactly.2012-09-10
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For the geometric description, observe that all four vectors are linear combinations of $(1,-1,0)$ and $(0,0,1)$. So the span is the plane that contains $(0,0,0)$ and those two points. This plane includes the $Z$-axis, and makes $45^\circ$ angles with the $X$- and $Y$-axes.