according to the definition of linear dependency vectors $v_1,...,v_n$ are linearly independent $iff$ $c_1v_1$+$c_2v_2$+$...$+$c_nv_n≠0$. One can also do the gaussian elemination to get which columns are independent but i'm trying now the first method.
Let $A=\left[\begin{array}{rrrr} c_1 & c_2 & c_3 \\ 1 & 2 & 1 \\ 0 & 3 & 2 \\ 4 & 2 & 0 \\ 0 & 3 & 2 \end{array}\right]= \left[\begin{array}{r}0\\0\\0\\0\end{array}\right]$
The book states that ${c_1-2c_2+3c_3=0}$, meaning that they are not linearly independent. How did they come out with the result, and namely how can one calculate it ? My head is going to blow! I cannot find a way to calculate it.
($c_1$ is the 1$^{st}$ column , $c_2$ the 2$^{nd}$..)
$\endgroup$ 24 Answers
$\begingroup$A matrix $A$ has linearly independent columns if and only if $A$ has a trivial nullspace. Indeed, if $$ A=\begin{bmatrix} \vec c_1 & \dotsb & \vec c_n \end{bmatrix} $$ then the equation $A\vec x=\vec 0$ is equivalent to $$ x_1\vec c_1+\dotsb+x_n\vec c_n=\vec 0 $$ Now, for your $A$ note that $$ \DeclareMathOperator{rref}{rref}\rref \begin{bmatrix} 1 & 2 & 1 \\ 0 & 3 & 2 \\ 4 & 2 & 0 \\ 0 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1/3\\ 0 & 1 & 2/3 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} $$ This tells us that $A\vec x=\vec 0$ if and only if \begin{align*} x_1 &= \frac{1}{3}\,x_3 & x_2 &= -\frac{2}{3}\,x_3 \end{align*} Taking $x_3=3$ then gives your result.
Note, however that this linear combination is not unique.
$\endgroup$ 0 $\begingroup$Let's try working out what $c_1-2c_2+3c_3$ is.
$$\begin{bmatrix} 1 \\ 0 \\ 4 \\ 0\end{bmatrix}-2\begin{bmatrix} 2 \\ 3 \\ 2 \\ 3\end{bmatrix}+3\begin{bmatrix} 1 \\ 2 \\ 0 \\ 2\end{bmatrix}= \begin{bmatrix} 1-2(2)+3(1) \\ 0-2(3)+3(2) \\ 4-2(2)+3(0) \\ 0-2(3)+3(2) \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
How did we find $c_1-2c_2+3c_3$? Let's write $\alpha c_1 +\beta c_2 + \gamma c_3 = 0$. Then we get:
$$\alpha+2\beta+\gamma=0$$ $$3\beta+2\gamma=0$$ $$4\alpha + 2\beta = 0$$
Subtracting the third equation from the first, we get $-3\alpha+\gamma=0$. Now just pick a value for your $\alpha$ (say $1$), and you find that $\gamma=3$. Then substituting into the second equation, you find $\beta=-2$
$\endgroup$ 5 $\begingroup$You are missing an "for every $c_1, c_2,... \in R$" This means, that every solution which gives you zero vector proofs dependency.
And one solution is $v_1-2v_2+3v_3=0$, the $c$s are the scalars.
How to calculate that: The second row tells you, that $3 c_2 +2c_3 =0$ and the third $4c_1+2c_2=0$. So just pick one of the $c$s and calculate the rest. Hint: try $c_1=1$.:)
$\endgroup$ 2 $\begingroup$Since we have that rank $A = 2$ (every $3\times 3$ determinant is equal to zero and there exists at least one $2 \times 2$ determinant not equal to zero), it is clear that the vectors $c_1, c_2, c_3$ are linearly dependent. Thus, we have: $$\begin{array}[t]{l}c_1 = a c_2 + b c_3\\\\ \begin{bmatrix} 1 \\ 0 \\ 4 \\0 \end{bmatrix} = \begin{bmatrix} 2a \\ 3a \\ 2a \\ 3a\end{bmatrix} +\begin{bmatrix} b \\ 2b \\ 0 \\2b \end{bmatrix} \\ \end{array}$$ So:
$$\left\{\begin{array}{l} 2a = 4\\[2ex] 3a + 2b = 0 \end{array}\right. \implies a = 2 \text{ and } b = -3,$$ which means: $c_1 = 2c_2 - 3c_2$.
$\endgroup$
