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Problem: Let $\vec{v}=(5,9), \ \vec{u}=(3,-2)$ and $\vec{w}=(2,1)$. Determine the nature of their linear dependency.

Attempt: So, we are looking for constants $k_1,k_2,k_3$ such that $k_1\vec{v}+k_2\vec{u}+k_3\vec{w}=0.$ We can write this as

$$k_1\left[\begin{matrix} 5 \\ 9 \end{matrix}\right]+k_2\left[\begin{matrix} 3 \\ -2 \end{matrix}\right]+k_3\left[\begin{matrix} 2 \\ 1 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$

which in canonical form is a system of linear equations in terms of $k_1,k_2$ and $k_3$.

$$ M\vec{k}=\left[ {\begin{array}{cc} 5 & 3 & 1 \\ 9 & -2 & 1 \\ \end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right].$$

Row echelon form on $M$ gives

$$ \left[ {\begin{array}{cc} 5 & 3 & 1 \\ 0 & -\frac{37}{5} & -\frac{4}{5} \\ \end{array} } \right],$$

This means that there are infinite solutions, but in order for them to be linearly independant, there should only exist one unique solution. Thus we have shown that the vectors $\vec{v},\vec{u},\vec{w}$ are linearly dependant. This means that $\text{span}\{\vec{v},\vec{u},\vec{w}\}=\mathbb{R}^2,$ because the addition of one of the vectors does not add another dimension to the span of the other two. The third vector that is linearly dependant of the other two, lies in their span.

Have I understood the concept of linear dependancy and span correctly? Any constructive input is very welcome!

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    $\begingroup$ Note that at most $n$ vectors can be linearly dependant in $\mathbb R^n$, so for any $3$ vectors in $\mathbb R^2$, it is enough to check if they are pairwise dependant. $\endgroup$ – George Sep 20 '17 at 14:03
  • $\begingroup$ There is an almost identical example(s) provided here en.wikipedia.org/wiki/Linear_independence $\endgroup$ – AmorFati Sep 20 '17 at 14:04
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Yes, you understood correctly the concept of linear dependency. And you proof is correct too.

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You have made a mistake, it should be:

$$M\vec{k}=\left[ {\begin{array}{cc} 5 & 3 & 2 \\ 9 & -2 & 1 \\ \end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$

The solutions are of the form $\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right]$ for $\alpha \in \mathbb{R}$.

Thus, $$-7\vec{v}-13\vec{u}+37\vec{w} = 0$$

so $\{\vec{v}, \vec{u}, \vec{w}\}$ is linearly dependent in $\mathbb{R}^2$.

In fact, any three distinct vectors in a two-dimensional space must be linearly dependent.

Edit:

I found the solutions by reducing the matrix to row echelon form: $$\left[ {\begin{array}{cc} 5 & 3 & 2 \\ 9 & -2 & 1 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 5 & 3 & 2 \\ -1 & -8 & -3 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 8 & 3 \\ 5 & 3 & 2 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 8 & 3 \\ 0 & -37 & -13 \\ \end{array} } \right] \sim \left[ {\begin{array}{cc} 1 & 0 & \frac{7}{37} \\ 0 & 1 & \frac{13}{37} \\ \end{array} } \right]$$

The first row gives $k_1 = -\frac{7}{37}k_3$, and the second row gives $k_2 = -\frac{13}{37}k_3$.

Thus, the solutions are given by $k_3\cdot \left[\begin{matrix} -\frac{7}{37} \\ -\frac{13}{37} \\ 1 \end{matrix}\right]$ for $k_3 \in \mathbb{R}$, which we can also write as $$\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right] \text{ for } \alpha\in\mathbb{R}$$ by setting $k_3 = 37\alpha$.

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  • $\begingroup$ How did you get the second line? the solutions are of the form...?` $\endgroup$ – Parseval Sep 20 '17 at 14:14
  • $\begingroup$ @Parseval I have added it. Just reduce the matrix using row transformations to something more sparse. $\endgroup$ – mechanodroid Sep 20 '17 at 14:32
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multiplying your second equation by $-2$ and addin g to the first we get $$-13k_1+7k_2=0$$ we get $$k_1=\frac{7}{13}k_2$$ therefore your vectors are dependent

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  • $\begingroup$ How do you conclude this based on your last formula? You're saying that $k_2$ is $k_1$ scaled by a factor of $\frac{7}{13}.$ How does this show that the vectors are dependant? $\endgroup$ – Parseval Sep 20 '17 at 14:16
  • $\begingroup$ there exist other Solutions the $$k_1=k_2=k_3=0$$ choose for instance $$k_2=1,k_1=\frac{7}{13}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 20 '17 at 14:18
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You worked well, but there's a little glitch.

The fact that the vectors are linearly dependent does not imply their span is $\mathbb{R}^2$: consider $(1,0)$, $(2,0)$ and $(3,0)$ to see why.

On the other hand, your matrix $M$ has two pivot columns, so indeed the span has dimension $2$ and therefore the three vector you were given do span $\mathbb{R}^2$.

For just establishing whether the three vectors are linearly dependent, it suffices to consider the fact that they're three and the space has dimension $2$: a set of $m$ vectors in a space of dimension $n$ is linearly dependent as soon as $m>n$; if $m\le n$ it can be linearly dependent or independent.

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