Linear independent property for vectors with entries that can go to infinity

Question

I have a non-linear controlled dynamical system for which the state $\begin{bmatrix}{x_1\\x_2\\x_3}\end{bmatrix}$ lives in $\chi=\mathbb{R} \times \mathbb{S}^1 \times \mathbb{S}^1$ and I have to prove the global controllabily. Using the Lie Brackets theory I have found the following three vectors (stucked in the coloumn of the following matrix): $\begin{bmatrix}{cos(x_2) \quad sec^2(x_3)sin(x_2) \quad \quad 0\\ sin(x_2) \quad -sec^2(x_3)cos(x_2) \quad 0 \\ tan(x_3) \quad \quad \quad \quad 0 \quad \quad \quad \quad sec^2(x_3)} \end{bmatrix}$ and I want to prove that they are linearly indipendent. However I have noticed that $x_3$ can vary in $[0,2\pi]$ and if $x_3=\pi /2$, the $sec^2(x_3)$ goes to infinity. Can I say also in that case that vectors are linearly indipendent?

Answer 1

By the way, I notice $x_1$ is missing from your matrix, I presume intentionally.

If you aim to show the three columns are independent for all values of $x_2$ and $x_3$, you cannot because the matrix is undefined when $x_3 = \pi/2 $ or $3\pi/2$.

Otherwise, you can just calculate the determinant. You will find with a little manipulation that it is $-\sec^4(x_3)$ and that is always non-zero when it is defined, so linear independence follows.