What is the fastest way of calculating whether vectors are linearly independent?

Question

Is the following family $(u_1, u_2, u_3)$ linearly independent? $$ u_1 := \begin{pmatrix} 1 \\ 3 \\ 5 \\ -1 \end{pmatrix} , u_2 := \begin{pmatrix} 1 \\ -1 \\ -3 \\ 3 \end{pmatrix}, u_3 := \begin{pmatrix} 3 \\ 2 \\ 1 \\ 4 \end{pmatrix}$$

So, do I have to form a matrix $U$ with columns $u_1$, $u_2$ and $u_3$ and use gaussian elimination to see whether in every column is a pivot or is there a faster way? The problem is meant to be solved fast, that is why I am asking. I have the feeling that I am missing some important information.

Answer 1

I'm not sure this is any faster, but here's another approach.

Clearly, no two of these vectors are linearly dependent. Thus, if these three vectors are linearly dependent, then $u_3$ must be a linear combination of $u_1$ and $u_2$. Suppose $u_3 = \alpha u_1 + \beta u_2$. Looking at the first entry, we get $\alpha + \beta = 3$. Looking at the second entry, we get $3\alpha - \beta = 2$. Together, these two equations imply that $\alpha = 5/4$ and $\beta = 7/4$. Using these values for $\alpha$ and $\beta$, we confirm that $5\alpha - 3\beta = 1$ and $-\alpha + 3\beta = 4$. Therefore,

$$u_3 = \frac{5}{4} u_1 + \frac{7}{4} u_2$$

so these three vectors are linearly dependent.

Answer 2

With Gaussian elimination, it is not very long to prove the rank is $2$: \begin{align} \begin{bmatrix} 1&1&3\\ 3&-1&2\\ 5&-3&1\\-1&3&4 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&1\\-1&2&3\\-3&1&5\\3&4&-1 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&1\\ 0& 5 & 4\\-3&1&5\\ 0 & 5& 4 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&1\\ 0& 5 & 4\\ 0&10&8\\ 0 & 0& 0 \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1&3&1\\ 0& 5 & 4\\ 0&0&0 \\ 0 & 0& 0 \end{bmatrix} \end{align}

Answer 3

Consider $u_2-u_1 = \begin{pmatrix}0\\-4\\-8\\4\end{pmatrix}$ and $u_3-3u_1 = \begin{pmatrix}0\\-7\\-14\\7\end{pmatrix}$.

These two vectors are linearly dependent. Therefore $u_1,u_2,u_3$ are linearly dependent.