What is the fastest way of calculating whether vectors are linearly independent? 
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.
 A: 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.
A: 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}
A: 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.
A: Arrange the Vectors as columns of Matrix $A$. Find the Metric Tensor $A^TA$
(or $A^{\dagger}A$ if the Vectors have complex numbers). If the Determinant
$|A^TA|\neq0$ (or $|A^{\dagger}A|\neq0$) then the Vectors are Linearly Independent. Otherwise they are Linearly Dependent.
To learn more about Metric Tensor please check
Gramian Matrix / Gram Matrix / Metric Tensor
Gram Matrix / Metric Tensor Calculator
