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$^{nd}$..)
Best Answer
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.