Rank of a set of vector fields

Question

Let $$D= {\mathbf R}^3$$

$$f_1=\begin{bmatrix}2x_2\\1\\0\end{bmatrix}$$ $$f_2=\begin{bmatrix}1\\0\\x_2\end{bmatrix}$$

How do I calculate the $$rank\{f_1,f_2\}$$

The answer is "2" for all x in D. But I dont know how this answer was arrived at.

Im new to this area, and any help is much appreciated

Answer 1

Whatever $x_2$ may be, $f_1$ and $f_2$ are linearly independent (because none of them is a multiple of the other one). Therefore, they span a space with dimension $2$. In other words, $\operatorname{rank}\langle f_1,f_2\rangle=2$.

Answer 2

It is to show that for all $x_2$ the vectors $f_1,f_2$ are linearly independent. To this end show that $s,t \in \mathbb R$ and $0=sf_1+tf_2$ imply that $s=t=0$.

Answer 3

Let note that for any $x_2$ we have $f_1,f_2\neq 0$ and consider

$$af_1+bf_2=0$$

then

• $2ax_2+b=0$

• $a=0$

• $bx_2=0\implies b=0$

which has solution if and only if $a=b=0$ that is precisely the definition for $f_1$ and $f_2$ to be linearly independent and therefore $rank\{f_1,f_2\}$.