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Determine if the following set of vectors is linearly independent:

$$\left[\begin{array}{r}2\\2\\0\end{array}\right],\left[\begin{array}{r}1\\-1\\1\end{array}\right],\left[\begin{array}{r}4\\2\\-2\end{array}\right]$$

I've done the following system of equations, and I think I did it right... It's been such a long time since I did this sort of thing...

Assume the following: \begin{equation*} a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*} Determine if $a=b=c=0$: \begin{align} 2a+b+4c&=0&&(1)\\ 2a-b+2c&=0&&(2)\\ b-2c&=0&&(3) \end{align} Subtract (2) from (1): \begin{align} b+c&=0&&(4)\\ b-2c&=0&&(5) \end{align} Substitute (5) into (4): \begin{equation} c=0 \end{equation}

So now what do I do with this fact? I'm tempted to say that only $c=0$, and $a$ and $b$ can be something else... but I don't trust that my intuition is right.

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If $c=0$ then you must have $b=0$ and then you must have $a=0$. Hence they are linearly independent. –  copper.hat Jun 6 '13 at 1:42
    
From $c=0$ and $b-2c=0$ you can conclude? And then what about $a$? You were doing fine. The same thing, with less writing, can be done using row reduction. –  André Nicolas Jun 6 '13 at 1:42
    
substitute $c=0$ back into (4) or (5) to show that $b=0$ and then both $b=0$ and $c=0$ into (1) or (2) to show that $a=0$. By definition they are then linearly independent. –  Tpofofn Jun 6 '13 at 1:43
    
@AndréNicolas I'm only starting to learn about matrices now... this was taught in this method so I presume this is how I have to do it on the assignment. –  agent154 Jun 6 '13 at 1:50
    
Yes, at the beginning it makes sense do do things directly from the definition. –  André Nicolas Jun 6 '13 at 1:51

2 Answers 2

up vote 5 down vote accepted

You just stopped too early:

You need to simply substitute $c = 0$ back into the two equations: from equation $(3)$, $c = 0 \implies b = 0$.

With $b = 0, c = 0$ substituted into equation $(1)$ or $(2)$, $b = c = 0 \implies a = 0$. So in the end, since

$$\begin{equation*} a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*}\implies a = b = c = 0$$ the vectors are linearly independent.

You could have, similarly, constructed a $3\times 3$ matrix $M$ with the three given vectors as its columns, and computed the determinant of $M$: We know that if $\det M \neq 0$, the given vectors are linearly independent.

$$M = \begin{bmatrix} 2 & 1 & 4 \\ 2 & -1 & 2 \\ 0 & 1 & -2 \end{bmatrix}$$

$$\det M = 12 \neq 0 \implies \;\;\text{linear independence of columns}$$

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you can take the vectors to form a matrix and check its determinant. If the determinant is non zero, then the vectors are linearly independent. Otherwise, they are linearly dependent.

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